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A 































MODERN 

HIGH SCHOOL ALGEBRA 


BY 

WEBSTER WELLS, S.B. 

M 

AUTHOR OF A SERIES OF TEXTS ON MATHEMATICS 

AND 

WALTER W. HART, A.B. 

ASSISTANT PROFESSOR OF MATHEMATICS, UNIVERSITY OF WISCONSIN 
COURSE FOR THE TRAINING OF TEACHERS 


J „ * 

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D. C. HEATH & COMPANY 

BOSTON NEW YORK CHICAGO 




Copyright, 1923, 

By D. C. Heath & Co. 

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PRINTED IN U.S.A. 


PREFACE 


The most notable undertaking in the field of instruction in mathe¬ 
matics in the past decade has been the work of the National Committee 
on Mathematical Requirements, culminating in its summary report, 
recently issued, and in its final report which will appear in a short time. 
One immediate result of this undertaking has been a revision of the 
definitions of the units in high school mathematics by the College En¬ 
trance Requirements Board. The extent to which the Wells and Hart 
High School Algebra, written eleven years ago, harmonizes with these 
recent reports is most gratifying. 

This Modem High School Algebra incorporates, first of all, the changes 
suggested by the eleven added years of experience of its author in teach¬ 
ing the subject to boys and girls of high school grade; and secondly, 
other changes which are suggested by these recent reports. 

It is a new book. While, naturally, it is based upon the Wells and 
Hart High School Algebra, the text has been rewritten wherever any 
improvement could be made in the instruction or expression. The 
explanations in the text are often more complete than those of the 
High School Algebra, — and those represented a high-water mark eleven 
years ago. This is done to help teachers — not to endeavor to super¬ 
sede them. The examples are all new, and so, too, are almost all the 
problems. Enough examples and problems are given in the text so that 
teachers will not need to seek drill material from other sources. 

The new features of this text are : 

1. Increased emphasis on the formula. 

In Chapter I, the formula is made the means of transition from 
arithmetic to algebra. It is emphasized throughout the text as 
its importance demands. (See pages 28, 61, 115, 242, etc.) 

2. Abundant use of geometrical material both as a means of instruc¬ 
tion and as problem material. 

The geometrical concepts and facts are taught inductively. 
This is intuitive geometry. Only those geometrical facts have 
been used which can be unified with the algebra. This is unified 
general mathematics — not mere mixed mathematics. (See 
pages 4-8, 22-25, 61-62, 76-78, 122, 442, etc.) 

iii 



IV 


PREFACE 


3. The function concept has been foreshadowed in various forms in 
the text, — and that seems about all that experience with beginning 
pupils warrants. 

Besides the obvious contribution to this concept of the work 
with the formula and the graph, see the development of the idea 
of dependence or relationship as in Examples 4-7, pages 2-3; § 23 ; 
page 241; page 282; page 406. 

4. The elimination from the suggested minimum course for the first 
year of H. C. F. (page 153); “ signs in fractions ” (page 181); the theory 
of proportion, which belongs more properly in a course in geometry 
(pages 203-207); literal equations (pages 198, 241, 278); linear 
equations in \/x and 1 /y (page 243); sets of equations having three 
unknowns (page 245); operations with surd expressions (page 257); 
simultaneous quadratics (page 279); imaginaries (page 285); gen¬ 
eralized forms of special products, and factoring (page 288). 

These topics appear in sections marked supplementary topics. Omis¬ 
sion of them leaves a minimum course which every school can master 
in a single year. This minimum course contains all the topics in al¬ 
gebra recommended by the National Committee and by the College 
Entrance Requirements Board. It is the briefest minimum course in al¬ 
gebra that has been proposed for the first course. The supplementary 
sections provide material needed for certain local courses of study and 
for teachers who need extra material for especially bright pupils. 

The author cannot agree with the National Committee on the 
desirability or necessity of omitting long division of polynomials 
and square root of polynomials. These do not present any dif¬ 
ficulties for pupils, and they are good teaching material. 

Similarly, in the part of the text designed for the third semester of 
algebra, certain sections are designated as supplementary. 

5. A brief chapter on the trigonometry of the right triangle is in¬ 
cluded. (See page 416.) 

6. The consistent use of “ charts ” in the solution of problems is 
suggested. 

These charts have been found a great help to pupils who are 
learning to solve problems (pages 85, 87, 88, etc.). 

7 . Lists of classified examples and problems at the back of the text 
for special drill purposes, beginning on page 426. 


PREFACE 


V 


8. Timed tests on fundamentals on page 451. These tests will 
enable those who are using this text to compare results in an interest¬ 
ing way with others who are using the same text. 

9. Summary of words and processes which the pupils should know 
at the end of the course. (See page 460.) 

The following features of this text have met with distinct approval 
for a decade by users of Wells and Hart texts, and are sanctioned either 
explicitly or implicitly by the report of the National Committee. 

1. Each topic taught is used in the solution of equations. 

This gives a motive for the study of each topic, permits grada¬ 
tion of material, and emphasizes the equation. (See §§ 14, 43, 
47, 55, etc.) 

2. Problems receive special attention. 

The different kinds are introduced gradually, and are taught 
carefully when they first appear. (See pages 85, 87, 101, 145, 
etc.) They appear repeatedly thereafter in miscellaneous lists. 

3. Factoring is cut down to its simplest possible form by omitting 
from the first treatment all types not used in the first course in algebra. 

The omitted types are taught in Chapter XV. This chapter 
provides a desirable review of factoring from a higher point of view, 
early in the third semester of algebra. Teachers will find that 
study of this chapter late in the course will prove effective prep¬ 
aration for any required examination. 3 

4. Thoughtful solutions of equations and examples are encouraged 
by subordinating “ transposition,” “ clearing of fractions,” etc. 

Observe the omission of these processes until page 104 and the 
suggestion in § 79 that they can be omitted altogether. Notice 
the convenient symbols A, S, M, and D, used in solving equations. 
(See pages 48, 59, etc.) 

5. Attempt is made to maintain and increase skill in arithmetic 

by emphasizing evaluation (pages 2, 26); by checking by substitution; 
by the use of fractional and decimal coefficients (pages 8, 9, etc.); by 
expressing roots decimally (page 261). 

6. Graphical work is taught and used where it is most helpful, just 
before simultaneous equations. Thereafter it is used whenever it aids 
in giving instruction in some algebraic process. (See pages 200-224; 
282-287; 307; 309-315.) 

March, 1923. 

























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% 



















CONTENTS 


I. Literal Numbers. 

II. Positive and Negative Numbers 

Addition of Positive and Negative Numbers 
Multiplication of Positive and Negative Numbers 

III. Addition and Subtraction of Polynomials 

Addition of Monomials .... 

Addition of Polynomials .... 

Subtraction. 

Subtraction of Positive and Negative Numbers 
Subtraction of Polynomials .... 

IV. Parentheses . . . 

Removal of Parentheses .... 

Introduction of Parentheses .... 

V. Multiplication. 

Multiplication of Monomials 
Multiplication of Polynomials by Monomials 
Multiplication of a Polynomial by a Polynomial 

VI. Division. 

Division of a Monomial by a Monomial 
Division of Polynomials by Monomials 
Division of Polynomials by Polynomials 

VII. Simple Equations. 

Properties of Equations .... 

VIII. Special Products and Factoring 
Quadratic Equations by Factoring 
vii 


PAGE 

1 

. 31 

. 35 

. 37 

. 41 

. 42 

. 45 

. 49 

. 51 

. 53 

. 63 

. 63 

. 66 

. 74 

. 75 

. 77 

. 80 

. 93 

. 96 

. 98 

. 99 

. 103 
. 104 

. 118 
. 147 





Vlll 


CONTENTS 


IX. Fractions. 

Reduction of Fractions 
Multiplication of Fractions 
Division of Fractions 
Addition and Subtraction of Fractions 
Complex Fractions .... 


156 

156 

160 

163 

165 

179 


X. Simple Fractional Equations . 
Ratio and Proportion 

XI. Graphical Representation 


. 186 
. 203 

. 208 


XII. Simultaneous Linear Equations . . 

Literal Simultaneous Equations (Supplementary) 
Equations containing Three Variables (Supp’y) 


225 

241 

245 


XIII. Square Root and Quadratic Surds 
Quadratic Surds (Supplementary) 


248 

257 


XIV. Quadratic Equations .260 

Complete Quadratic Equations.265 

XV. Special Products and Factoring (Advanced-Topics) 288 

Remainder Theorem (Supplementary) . . . 298 

Synthetic Division (Supplementary) .... 299 

Factor Theorem (Supplementary) .... 301 


XVI. Quadratic Equations Having Two Variables. 

Graphical Solution.309 

XVII. Systems of Equations Involving Quadratics . 316 

XVIII. The Theory of Quadratic Equations . . . 326 

XIX. Literal, Fractional, and Negative Exponents . 332 


XX. Radicals .343 

Imaginary Numbers.355 

Irrational Equations.359 

XXL Logarithms .363 






CONTENTS 


IX 


PAGE 

XXII. Progressions.380 

Arithmetic Progression.380 

Geometric Progression.388 

XXIII. The Binomial Theorem.397 

XXIV. Variation.406 

XXV. Determinants. 411 

XXVI. Trigonometry of the Right Triangle . . . 416 

Supplementary Exercises.426 

Timed Tests.451 

Definitions, Rules, and Processes to be Learned 460 

Index.464 








To the Teacher: 

This text is designed to aid you in promoting mastery of the 
beginnings of algebra by your pupils. Your attention is di¬ 
rected to the minimum course which remains when the supple¬ 
mentary topics are omitted. This minimum course is supple¬ 
mented by other lists of drill material at the back of the book, 
by timed tests, and by a list of words and processes which pupils 
should know at the end of the year. Use this material as means 
of mastering the minimum course. Insist upon mastery of 
it. Some pupils can accomplish this by working a few ex¬ 
amples ; others will need to work many examples. Some can 
do it quickly; others will require more time. Have your 
pupils check off in the list at the back of the book each new 
word and process as you encounter it in your course. 

The supplementary material should be considered optional. 
If necessary, agree that only certain types of problems shall be 
mastered by those who get the lowest passing mark. Let the 
higher marks be regarded as rewards for mastering other types 
of problems and for mastering some of the supplementary 
material. This point of view of mastery is discussed most fruit¬ 
fully by Professor Morrison, of the University of Chicago, in 
the School Review of January, February, and March, 1921. 

To the Pupil: 

Let your motto be: “I can do it.” 


x 




ALGEBRA 


I. LITERAL NUMBERS —FORMULA: 


1. Letters are used to represent numbers, in order to abbre¬ 
viate rules. Thug) P = BxR 


expresses the rule 

the 'percentage equals the base multiplied by the rate. 

B represents the base, R the rate per cent, and P the percentage. 


If B = 2500, and R = 5%, 
P = 2500 X .05 = 125. 


A letter used to represent a number is called a literal number. 

The use of literal numbers is one characteristic of algebra. 

2. Signs of multiplication. The symbol X is not used much 
in algebra as the sign of multiplication because it is like the 
letter x. A dot, *, placed above the line between two numbers 
means that the numbers are to be multiplied together. 

Thus, 6 • 7 means 6 times 7, or 42. 
a * b means a times b. 


Even the dot is omitted when one or both of the numbers are 
literal numbers. 

Thus, 6 a means 6 times a. It is read “six a .” 
xt means x times t. It is read “ex tee.” 


Historical Note. — The symbol X was first used by an English¬ 
man, Oughtred, about 1631. The symbol • was introduced by Leib¬ 
nitz in 1693. Multiplication was indicated in Hindu and Italian books 
by writing the factors side by side, as early as the thirteenth century. 

Look up the biographies of such men when they are mentioned in 
the historical notes. 


1 


2 


ALGEBRA 


3. The product of two or more numbers is the result obtained 
by multiplying together the numbers; the numbers themselves 
are the factors of the product. Every factor of a product is an 
exact divisor of the product. 

Thus, 5 and 8 are factors of 40. 


EXERCISE 1 


1. What does 5 m mean? 8 x? 7 t ? 

2. How much is, or what is the value of 10 x when x is 7 ? 
Also when a: is 13 ? When x is 2.5 ? 


3. What is the value of 20 y when y is 3 ? 
When y is ^ ? 


4 . Make a table like the one at the right, or 
complete this table orally. Opposite each value 
of x given in the first column, write or give the 
value of 6 Z in the second column. 


When y is .5? 


When 

Then 

x is 

6 z is 

3 


8 


11 


i 

s 

.5 

2 


Thus, opposite | we write 2, because when x is 6 x is 6 X or 2. 


6. See how rapidly you can complete the following tables: 


ah c 


When 

Then 

When 

Then 

When 

Then 

y is 

8 y is 

m is 

24 m is 

t is 

18 t is 

7 


i 

"3 


5 


12 


3 

T 


9 


2.5 


5 

¥ 


.6 


i 


7 

¥ 


i 


5 

"8 


11 

12 


2 

¥ 



6. a. What are the values of 9 y when y is 3, 5, 6, 10, 20 ? 
6. As y increases (becomes greater), what happens to 9 2/? 













LITERAL NUMBERS — FORMULAS 


3 


7. a. What are the values of 10 A when A is 15, 12, 10, 9, 4? 

b. As A decreases (becomes smaller), what happens to 10 A ? 

8. Suppose that a represents the length of A a B 

segment AB. What does 3 a then represent ? - 

9. a. How much do 5 doz. eggs cost at 40per dozen ? At 
50^ per dozen ? At x cents per dozen ? 

b. How much is 5 x cents when x is 60 ? 

10. A certain number is represented by n. What repre¬ 
sents the number 4 times as large? 11 times as large? 

11. If b is the number of feet in the base of a rectangle, and 
5 is the number in the altitude, what is the area ? 

12. If h is the number of feet in the altitude of a rectangle, 
and if the base is 3 times as long, how many feet are there in 
the base ? 

13. a. What is the interest on $100 for 1 yr. at 5%? for 
3 yr. ? 

b. What is the interest on P dollars for 1 yr. at 5 % ? for 

3 yr.? 

14. a. An article cost $5. It was sold so that the profit was 
10 % of the cost. What was the profit ? 

b. An article cost C dollars. It also was sold so that the 
profit was 10 % of the cost. What was the profit ? 

c. In part 6, how much would this profit be if C = 20 ? 

15. a. How much do 8 tons of coal cost at $12 per ton ? 

b. How much do 8 tons of coal cost at p dollars per ton ? 

c. What is the value of the result in part b when p = 10 ? 

4. A formula is a rule of computation expressed by means 
of arithmetical and literal numbers, connected by mathematical 
signs w T hich tell what must be done with the numbers. 

Thus, in § 1, P = BR is a formula. It is called the percentage 
formula. 



4 


ALGEBRA 


The use of literal numbers in formulae is one of their most 
important uses. 

jy g 

5. Making and using some important - r 

formulae. a 

a. The formula for the area of a 
rectangle. E b F 


You know that 

The area of a rectangle equals the product of its base and 
altitude. 


Thus, if the base is 8 in., and the altitude is 15 in., the area is 8 X 15 
or 120 sq. in. 


If a 
b 

and A 
then 


the number of units of length in the altitude, 
the number of the same units in the base, 
the number of the corresponding surface units in 
the interior of the rectangle, 

A = ab. 


This formula for the area of a rectangle is used as follows: 
Example. — Find the area of the rectangle whose base is 2-J 
ft. and whose altitude is 15 in. 

Solution. — 1. The formula is A = ab. 

2. a = 15 in.; b = 2* ft. = 30 in.; A = ? 

3. .*. A = 15 X 30 = 450 sq. in. 

Note 1. — In Step 3, the symbol .\ means “therefore.” 

Note 2. — To “explain” Step 3, say, “I substituted 15 for a and 30 
for b in the formula.” “Substitute” means “put in place of.” 


EXERCISE 2 

Solve the following problems, using the form illustrated 
above. 

1. What is the area of the rectangle whose base is 13 ft. 
and whose altitude is 8 ft. ? 

2. Find the area of the rectangle whose base is 3 yd. and 
whose altitude is 4 ft. 




LITERAL NUMBERS — FORMULAE 


5 


3. Find the area of the rectangle whose base is 10 ft. 6 in. 
and whose altitude is 8 ft. 

b. The formula for the area of a parallelogram. 


x_w 

Z: l a u / 


The figure X YZW is a parallelogram. 

a. How does X YZW compare in area with 
FGHK ? 

b. If the altitude of the parallelogram is 5 
in. and the base is 12 in., what is the area of 
the rectangle FGHK ? What, then, must be 
the area of the parallelogram X YZW ? 

c. If the altitude of the parallelogram were 10 in. and the 
base 16 in., what would be its area? 


z 


d. Evidently the area of a parallelogram is the product of its 
base and altitude. 


.If a = the number of units of length in the altitude, 

b = the number of the same units of length in the base, 
and A = the number of corresponding surface units inside 
the parallelogram, 

then A = ab. 


c. The formula for the area of a triangle. 

The figure XYZ is a triangle. 

Notice that triangle XYZ incloses one 
half as much surface as parallelogram 
XYZW. 

Since the area of the parallelogram is 
the product of its base' and altitude, then 
the area of the triangle must be one half the product of its base 
and altitude. 

If the letters a, b , and A have the same meaning as above, 
then the formula for the area of a triangle is 










6 


ALGEBRA 


Example. — What is the area of a triangle whose base is 
yd. and whose altitude is 2 ft. ? 

Solution. — 1. The formula is A = \ ab. 

2. b = 2| yd. = 7 ft.; a = 2 ft.; A = ? 

3. Substituting in the formula, 

A = a • 7 • 2 = 7 sq. ft. 

EXERCISE 3 

1. Find the area of a parallelogram whose base is 18 ft. and 
whose altitude is 5 ft. 4 in. 

2. Find the area of a triangle whose base is 12^ ft. and whose 
altitude is 9 ft. 

3. Find the area of a triangle whose base is 3 ft. 4 in. and 
whose altitude is 15 in. (Express each in terms of inches or feet.) 

4. What is the area of the parallelogram whose base is 2j 
ft. and whose altitude is 20 in. ? 

5. The figure at the right is a rectangular parallelopiped. 
You know that its volume equals its length 
multiplied by its width, multiplied by its height, y/j. 

If the length, width, and height are represented j_. 
by l, w, and h, respectively, and the volume by 
V, then ^- 

V = Iwh 

is the formula for the volume of a rectangular parallelopiped. 

By this formula, find the volume of the rectangular parallelo¬ 
piped whose length is 9 in., width 12 in., and height 7 in. 

6. Find the volume of the rectangular 
parallelopiped whose length is 30 ft., width 
6 yd., and height 15 ft. 

7. The figure at the right is a pyramid. 

The volume of a pyramid is one third of 

the product of its base and altitude. If 
a = the number of units of length in the 















LITERAL NUMBERS — FORMULAE 


7 


altitude, b = the number of the corresponding surface units in 
the base, and V = the number of corresponding cubic units in 
the pyramid, then V = \ ab 

is the formula for the volume of a pyramid. 

Find the volume when the altitude is 12 in. and the base con¬ 
tains 10 sq. in. 

8. Find the volume of a pyramid whose altitude is 5 yd. 
and whose base contains 35 sq. ft. 

9. a. The formula for finding the simple interest on a sum 
of money is / = PRT. Express this formula in words. 

b. Find / when P = $3600, R = 5%, and T = 4 yr. 

10. Find the interest on $2500 invested at 6% for 3 yr. 

11 . Find the simple interest on $4600 invested at 8% for 2 yr. 
6 mo. 

12. a. The rule for finding the circumference of a circle is to 
multiply the radius by twice the number 3.1416, which is called 
pi. If r = the length of the radius, tt = the number 3.1416, 
and C = the circumference, this rule is expressed by the formula 

C = 2 7r r. 

6. Find the circumference of the circle whose radius is 5 in. 

13. 3iJ- instead of 3.1416 may be used as the value of 7r, wher¬ 
ever the number n r appears. Using this value of i r, find the 
circumference of the circle whose radius is 21 in. 

14. What is the length (circumference) of the circle whose 
diameter is 14 ft. ? 

16. What is the simple interest on $650 at 7% for 9 mo.? 
(Remember to express 9 mo. as a part of a year.) 

16. By the formula Z — 2 irrh, find Z when r = 14, h = 16, 
and tt — 34-. 

17. By the formula S — \ CL, find S when C = 32.4 and 
L = 15. 


8 


ALGEBRA 


18. By the formula S = rrL, find S when r = 28, L - 40, 
and 7r = 3^. 

19 . By the formula L = 2tt rh, find L when tt = 3.1416, 
r = 8, and h = 10. 

20. By the formula F = f C, find F when C = 45. 

6. Memorize the following formulae which you used in the 
preceding exercise. 


The formula for: 

a. The area of a rectangle is A = ab. 

b. The area of a parallelogram is A = ab. 

c. The area of a triangle is A = \ ab. 

d. The volume of a rectangular parallelopiped is V = Iwh. 

e. The volume of a pyramid is V = % ab, 

f. The circumference of a circle is C = 2 irr. 

g. Simple interest is J = PR T. 


7. An important law of multiplication. 

In arithmetic, 2x3x4 = (2x3)x4 = 6x4=24. 
Similarly, in algebra, 5 X 4 m = (5 X 4) X m = 20 m. 

Note. — The symbol ( ) is called 'parentheses. The numbers inside 
are to be combined before using them in any other addition, subtrac¬ 
tion, multiplication, etc. 

EXERCISE 4 


Give the following products orally, or write only the results. 


1. 7 

2. 3 

3. 9 

4. 8 


8 x 

11 b 

12 m 
15 t 


12-8 w 
11-7 t 
6 • 13 v 


8. 9 • 15 2 


9. 4 • f x 

10. 5 • f y 

11 . 6-fz 

12. 12 -±w 


13. 20 • f m 

14. 24 • | p 


15. 

16. 


36 

40 


17. a. One number is represented by 3 x. What represents 
the number twice as large? Five times as large? 
b. How much is each of these numbers when x is 2 ? 






LITERAL NUMBERS — FORMULAE 


9 


18. a . Let / represent the first of certain three numbers. If 
the second is 5 times the first, and the third is 3 times the second, 
represent each of thetn. 

b. Find the value of each of these numbers when / is 4. 

19. The area of a certain rectangle is 17 s square inches. 
What is the area of a parallelogram twice as large ? 

20. a. Edward is 3 y years of age. Will is 5 times as old. 
How old is Will? 

b. What is the age of each, if y is 2 ? 

8. An important law of division. 

Since 2 • 3 a = 6 a, then 6a -f 2 =3 a. 

Similarly, since 5 • 8 x = 40 x, then 40 x 4 
Also, 20 s -r- 5 = 4 s; 18 ?/ -=- 9 

EXERCISE 5 

Give orally or write only the quotients: 


1. 

25 t 

5 

5. 

36 x 

-r 36 

9. 

84 p 

12 

2. 

32 r -i- 

8 

6. 

29 y - 

r 29 

10. 

16 m 

-f- 16 

3. 

60 w -7 

-12 

7. 

39 t -r 

- 39 

11. 

.5 w ■ 

.5 

i. 

24 s -T- 

4 

8. 

2.5 z - 

^ 2.5 

12. 

i z * 

' i 

13. 

What i 

part of 32 

x is: 







a. 8 s? 6.16 s? c. 4s? d. s? 

14. What part of 28 t is : 

a. 2f? b. 7*? c. 4f? d . f? 

15 . a. By what must you divide 6 x to get s? 

b. By what must you divide 18 r to get 2r? 

c. By what must you divide 25 z to get z ? 

d. By what must you divide 2.5 x to get x ? 

e. By what must you divide .56 y to get y ? 

Historical Note. — The symbol -s- was introduced by John Pell, 

who lived during the seventeenth century. 


- 5 = 8 )^ 
= 2 y. 


10 


ALGEBRA 


9. An equation expresses an equality of numbers. A 

formula is one kind of equation. 

Thus, A = 3 b is an equation. 

It states that the number A is 3 times the number b. 

Again, 3 x + 11 = 26 is an equation. 

It states that the result of adding 11 to 3 times a certain number, x, 
is 26. 

The numbers on the right side of the equality sign form the 
right member or the right side of the equation; the ones on the 
left side of the equality sign form the left member or left side 
of the equation. 

10. Solving an equation. 

Consider the equation 3 x + 11 =26. 

When x = 5, 3 x + 11 is 3 • 5 + 11 or 15 + 11. 

This does indeed equal 26. 5 is said to satisfy the equation. 

5 is called a root of the equation. 

But when x is any other number, 3 x + 11 does not equal 26. 

Thus, if a; = 8,3-8 + 11 =24 + 11 = 35. 

In such an equation, the letter represents an unknown sfiumber; 
often for this reason, the letter itself is called the unknown 
number. 

EXERCISE 6 

1. Does 6 satisfy the equation 3 x + 8 = 26 ? 

2. Does 10 satisfy the equation 4 y + 7 = 47 ? 

3. Does 3 satisfy the equation 52 + 11 =31? 

4. Does 6 satisfy the equation 52 + 11 =31? 

5. What number does satisfy the equation 5 2 + 11 =31? 

11. Rule I used in solving equations. 

Suppose the weight and the sugar just balance on the scales 
pictured at the right. 

Then one half the sugar will also balance 
one half the weight. It is equally clear that 
one half the sugar will not balance one third of the weight. 











LITERAL NUMBERS — FORMULAE 


11 


Similarly, in an equation like 10 x = 80, 
if both sides are divided by 2, the results must be equal; 

thatis > 5 x = 40. 


But if one side is divided by 2, and the other side is divided 
by some other number, the results will not be equal. 

Rule. — Both members of an equation may be divided by 
the same number without destroying the equality. 

Example 1. — Solve the equation 17 y = 391. 

Solution. — 1. y can be got from 17 y by dividing 17 y 
by 17. Then, in order to keep the two sides equal, both sides 
must be divided by 17. 

2. Then, y = 23. 

In order to abbreviate the written explanations of solutions, 
the symbol D will be used as follows: 

Di 7 will mean “ divide both sides of the preceding equation 
by 17.” 

Example 2. — Solve the equation 6.5 w = 11.7. 


Solution. — 1. 6.5 w = 11.7. 

2. De.s w = 1.8. 


Note. — In Step 2, what does D 6 .5 mean? Notice that 6.5 w -s- 6.5 
= w ; and that 11.7 - 5 - 6.5 = 1.8. Also, that w = 1.8 by Rule I. 


1.8 
6?5.)11?7.0 
65 
520 
520 


23 

17)391 

34 

51 

51 


EXERCISE 7 


Solve the following equations, using the form of solution 
illustrated in Example 2. 


1. 11 x = 286 

2. 15 y = 285 

3. 17 z = 476 

4. 22 W = 814 

6. 9 A = 2583 


6. 14 m = 133 

7. 27 p = 221.4 

8. 35 r = 539 

9. 44 5 = 158.4 

10. 56 c = 420 


11 . 

12 . 

13. 

14. 

15. 


32 w = 528 
8.4 y = 168 
.12 m = 24 
1.35 x = 911.25 
.75 p = 1800 





12 


ALGEBRA 


Note. — Some of the problems which follow can be done by arith¬ 
metic alone. Do them algebraically even if the arithmetical solution 
appears easier to you at the present time. 

16. If a certain number be multiplied by 17, the product is 

144.5. What is the number? 

Solution. — 1. Let n = the number. 

2. Then 17 n = 17 times the number. 

3. Then 17 n = 144.5, according to the statement of the 

problem. 

Finish this problem by dividing both sides by 17. 

17. If a certain number be multiplied by 13, the product is 
247. What is the number? 

18. If a certain number be multiplied by 8.5, the result is 221. 
What is the number ? 

19. How many articles at $1.75 each can be bought for $40.25 ? 
Solution. — 1. Let n = the number of articles. 

2. n • $1.75 or 1.75 n = the cost of the articles. 

3. .*. 1.75 n = 40.25. 

(Complete this problem.) 

20. How many articles at $2.15 each can be bought for $38.70 ? 

21. What number, multiplied by 2.7, gives 9.45 as product ? 

22. How many miles per hour did an auto travel which went 
172 miles in 8 hours ? 

23. At what price must an article be purchased if 16 of them 
cost $13.60? 

24. If a certain number be multiplied by 12.5, the product 
is 217.25. What is the number ? 

25. How much must each of 25 persons contribute to make 
up a fund of $156.25, if all contribute equally ? 

26. The product of two numbers is 499.5. One of them is 

18.5. What is the other ? 

27. How much must a boy save each month to have $6.75 
at the end of 15 months ? 




LITERAL NUMBERS — FORMULAE 


13 


12. Rule II used in solving equations. 

Suppose the sugar S balances the weight W on the scales. 
Then double the amount of sugar will also 
balance double the amount of weight. 

It is equally clear that double the amount 
of sugar will not balance three times the amount of weight. 
These facts help you understand the 



Rule. — Both members of an equation may be multiplied by 
the same number without destroying the equality. 

Example 1.. — Solve the equation ■§ x = 36. 

Solution. — 1. f x = 36. 

2. We decide to get rid of the fraction f. If we multiply it by 8, 
then 8 X f is 3. 

Then, in order to keep the two sides equal, we multiply both sides 
of the equation by 8. 

3. Then, 3.3 = 288. 

4. D 3 x = 96. 


Check. 


3 12 

Does | X 90 = 36 ? Yes. 

p 


Note. — Observe that 8 • f x = 3 x ; that 8 • 36 = 288; that 3 x = 288 
by Rule II. All this is done mentally when passing from Step 2 to 
Step 3. 


Instead of writing in full 

“ multiply both sides of the previous equation by 8,” 
we shall write M 8 . 


Example 2. — Solve and check the equation 92.4 = % w. 
Solution. — 1. 92.4 = £ w. 

2. M 3 277.2 = 7 w, 

or 7 w = 277.2. 

3. D 7 w = 39.6. 

Check. — Substituting 39.6 for w in the original equation, 

13 2 

does 92.4 = ? X 3 W Yes. 

3 

Note. — What does M 3 mean in Step 2 above? D 7 in Step 3? 


13.2 

_7 

92.4 







14 


ALGEBRA 


EXERCISE 8 

Solve the following equations, using the form of solution 
illustrated in Example 2. 


1. 

f * = 

■- 40 

6. 


= 80 

11 

12 = 

: TE A 

2. 

iv = 

= 84 

7. 

ir- 

= 77 

12. 

36 = 

-t B 

3. 

TT Z 

= 54 

8. 

i s = 

= 120 

13. 

12/ : 

= 16 

4. 

A* : 

= 30 

9. 

fra 

= 189 

14. 

i * : 

= 28.7 

6. 

TI W 

= 9 

10. 

-hr c 

= 99 

16. 

fra 

= 50.75 


16. Fifteen ninths of a certain number is 75. What is the 
number ? 

Solution. — 1. Let n = the number. 

2. .'. hrn = 75. 

Complete the solution. 

17. Five ninths of a certain number is 55. What is the 
number ? 

18. Seven thirds of a certain number is 112. What is the 
number ? 

19. Eight fifths of the cost of a certain lot was $3000. What 
was the cost of the lot? 

20. Three eighths of the area of Lake Michigan is 8625 sq. 
mi. What is the area of Lake Michigan ? 

21. When asked her age, a girl replied, seven fifths of it is 21. 
What was her age ? 

22. Four sevenths of the members of a certain class were 
boys. If there were 28 boys, what was the total membership 
of the class? 

23. The selling price of a certain article was five thirds of its 
cost. What must have been its cost if it sold for $3.50? 

24. Five eighths of the cost of the Suez Canal was $62,500,000. 
What was the cost of the canal ? 



LITERAL NUMBERS — FORMULAE 


15 


25. Nineteen twentieths of the population of the United 
States in 1920 was 100,425,089. What was the population in 
that year ? 

13. A second important law of multiplication. 

You know that 2x3x4 = 6x4=24; 

that 3X2X4=6X4=24; 
that 2X4X3=8X3=24. 

That is, 2X3X4 = 3X2X4=2X4X3. 

Rule. — The factors of a product may be rearranged in any 
order before finding the product. 

This rule is used in algebra as follows : 

6-a*5 = 6*5‘a = 30a. 

3 ^ 3 

8 • 6 • - = B • 7 * 6 = 6 6. 

4 i 

The factors should be rearranged and multiplied mentally, 
however. 

EXERCISE 9 

Find mentally: 

1. 7 • m • 8 4. 12 • y • f 7. 2 • r • 5 • 3 

2. 6 • y • 9 5. 16 • x • £ 8. 6 • s • 4 • 5 

3. 7 • 2 • 3 6. 18 - 2 • | 9. 15 • t -8 

14. The second step in using formulae. 

You are now prepared to find any number in a simple formula 
like those given in § 6. This is a most valuable mathematical 
accomplishment. 

Example 1. — From the formula V = Iwh, find w if 
V = 1456, l = 7, and h = 16. 

Solution. — 1. The formula is V = Iwh. 

V = 1456; l = 7; h = 16; w = ? 


16 


ALGEBRA 


2. Substituting in the formula 

1456 = 7 • w • 16. 

3. 1456 = 112 w, 

or 112 to = 1456. 

4. Dm w = 13. 

Check. — Does 7 X 13 X 16 = 1456? Yes. 

i 


_ 13 

112)1456 

112 



336 


336 

13 

16 

7 

91 

91 

16 


144 

1456 


Example 2. — From the formula V = % hb find b when 
F = 117 and h = 13. 


Solution. — 1. Substituting in the formula 
117 = f • 13 • b, 

or ¥-b = 117. 

2. M 3 13 b = 351. 

3. D 13 b = 27. 

1 9 

Check. — Does ± X 13 X 27 = 117? Yes. 

3 


EXERCISE 10 

1. By the formula F = Iwh, find h when V = 672, l = 12, 
and w = 8. 

2. By the formula V = Iwh , find w when F = 2340, l = 20, 
and h = 13. 

3. By the formula V = Iwh , find Z when F = 1360, w = 8.5, 
and h = 10. 

4. By the formula A = Ic , find c when A = 3309.6 and 
l = 12 . 


6. By the formula 8=2 rrh, find h when S = 528, t = - 2 T 2 , 
and r = 14. 

6. By the formula A = \hb, find b when A = 189 and 
h = 14. 

7. By the formula F = ^ hb, find b when F = 247 and 
h = 19. 

8. By the formula F = %hb, find h when F = 232 and 
b = 29. 




LITERAL NUMBERS — FORMULAE 


17 


9. By the formula A = \ ab, find b when A = 47.5 and 
a 

10. By the formula C = 2 irr, find r when C = 66 if x = 

11. What is the altitude of the rectangle whose area is 594 sq. 
ft. and whose base is 33 ft. ? 

Solution. — 1. The formula for the area of a rectangle is 
A = ab. 

2. In this problem, A = 594; b = 33; a = ? 

Complete the solution as in Examples 1-10. 

12. What is the base of the triangle whose area is 651 sq. ft. 
and whose altitude is 21 ft. ? 

13. What is the altitude of the parallelogram whose area is 
153 sq. yd. and whose base is 8.5 yd. ? 

14. What is the altitude of the rectangular parallelopiped 
whose volume is 3240 cu. in., whose width is 15 in., and whose 
length is 24 in. ? 

15. What is the altitude of the pyramid whose volume is 396 
cu. in. and whose base is 27 sq. in. ? 

16. What is the base of the triangle whose area is 300 sq. ft., 
and whose altitude is 15 ft. ? 

17. What is the base of the parallelogram whose area is 372 
sq. ft., and whose altitude is 15.5 ft. ? 

18. What is the width of the rectangular parallelopiped 
whose volume is 5400 cu. in., whose altitude is 18 in., and 
whose length is 25 in. ? 

19. What is the length of the rectangular parallelopiped 
whose volume is 2550 cu. ft., whose width is 12£ ft., and whose 
height is 8-J ft. ? 

20. What is the base of a pyramid whose altitude is 16 in. 
and whose volume is 256 cu. in. ? 

Note. — Supplementary examples of the same kind are on page 426. 


18 


ALGEBRA 


15. Adding and subtracting numbers having a common factor. 

Any exact divisor of a number is a factor of it. 

Thus, 3 is a factor of 21. 

A common factor of two or more numbers is a factor of each 
of them. 

Thus, 3 is a common factor of 6 and 9. 

x is a common factor of 4 £ and 11 x. 

The result obtained by adding two or more numbers is called 
their sum. The numbers themselves are called addends. 

The method of adding numbers which have a common factor 
is taught in the following exercise. 


EXERCISE 11 

1. 3 times 7 + 2 times 7=5 times 7 

Because (7 + 7 + 7) + (7 + 7) = (7 + 7 + 7 + 7 + 7). 

2. 5 times 9 + 3 times 9=8 times 9 = ? 

3. (7 X 4) + (3 X 4) = ? X 4 = ? 

4. (2 X 6) + (3 X 6) + (8 X 6) = ? X 6 = ? 

5. 2*w + 3*?i + 8*7i = ? • w 

10. 10 x + 5 z + 3 x = ? 

11. 2y + 3y +y = 1 

12. 5m + 13m+m = ? 

13. 16 + + A + 9 A = ? 

14. Similarly, 8 times 3 — 6 times 3=2 times 3=6. 

15. (10 X 5) - (3 X 5) = ? X 5 = ? 

16. (15 X 9) - (11 X 9) = ? X 9 = ? 


6. 

7 t + 8t = U 

10. 

7. 

5 r + 6 r = ? 

11. 

8. 

Us + 5s = ? 

12. 

9. 

4 p + 15 p = ? 

13. 


17. 

11 t - 

3 t = H 

22. 

2 x + 6 x 

- Sx = ? 

18. 

16 r - 

II 

23. 

11 ?/ — 6 ?/ 

+ 9y = ? 

19. 

20 y - 

-112/ = ? 

24. 

15f + f- 

St = ? 

20. 

23 B 

- 18 B = ? 

25. 

22 s + 6 s 

- ^ = ? 

21 . 

36 W 

- 15 W = ? 

26. 

9 + 5g - 

3 g = ? 


LITERAL NUMBERS — FORMULAE 


19 


27. What is the perimeter of each of the following figures ? 



a be 


Obtain mentally the results for: 

28. | t + %t 31. 5 c+ .2 c 34. 6.25 t - 3.05 t 

29. f r - | r 32. 3 m + .35 m 35. 7\ w + 2£ w 

30. %y—ToV 33. 8.5 a: — 2.3 z 36. 5§ x — 3f x 

37. a. What is the perimeter of the square whose sides are 
m ft. long ? b. How much is the perimeter when m = 5 ? 

38. Answer the same questions if the figure is an equilateral 
triangle. 

39. a. The longer side of a parallelogram is 5 times as long 
as the shorter side. If the shorter side is n inches long, how long 
is the longer side ? 

b. What is the perimeter of the rectangle ? 

40. The length of a certain room is 6 a; ft. and its width is 
4 x ft. Its height is 9 ft. What is the total area of its four 
walls ? 

16. Addition and subtraction of literal numbers used in 
equations. 


Example. — Solve the equation 6 a + 9 a = 50. 


Solution. — 1 . 6 a + 9a = 50. 

2. Adding, 15 a = 50. 

3. D 15 a = 3!. 


10 

50 - 5-15 

3 

= 3i 


Check. — Substituting 3| for a in the original equation, 
Does 6 X 3^ + 9 X 3| = 50? 

Does 6 X - 1 # + 9 X = 50? 

Does 20 + 30 = 50? Yes. 









20 


ALGEBRA 


EXERCISE 12 


Solve and check the following equations : 


1. 

5 r + 3 r = 

96 

7. 

4 x + 8 x 

+ 6 x -- 

= 99 

2. 

Ilf + 4f = 

= 105 

8. 

18 t - 5 t 

-2 t = 

= 198 

3. 

oo 

1 

= 117 

9. 

5 c -f 9 c 

— 6 c = 

: 132 

4. 

17 y — 5 y 

= 180 

10. 

11 w — w 

+ 12 w 

= 77 

5. 

2.6 z + .9 z 

= 28 

11. 

25 z + z- 

- 8 z = 

96 

6. 

16.5 m — 4. 

2m = 123 

12. 

s + 15 s - 

-45 = 

75 


17 . Solving problems by means of equations is the second 
chief purpose of algebra. 

Example. — The sum of two numbers is 96. The larger 
number is 11 times the smaller. What are the numbers ? 

Solution. — 1. Let s = the smaller number. 

2. Then 11 s = the larger number. 

3. Then s + 11 s = the sum of these two numbers. 

4. /. s + 11 s = 96, since the sum of the numbers is 96. 

5. Adding, 12 s = 96. 

6. D 12 s = 8, the smaller number. 

7. .*. 11 s = 88, the larger number. 

Check. — Is the sum of 8 and 88 96? Yes. 

Is the larger, 88, eleven times the smaller, 8? Yes. 

Note. — There are other pairs of numbers whose sum is 96. Thus, 
16 and 80. But 80 is not eleven times 16. 

Also there are many pairs of numbers such that one is eleven times 
as large as the other. Thus, 5 and 55. But 5 + 55 is not 96. 

8 and 88 are the only two numbers whose sum is 96 and such that 
one of them is also 11 times the other. 

EXERCISE 13 

1. The greater of certain two numbers is 5 times the smaller. 
The sum of the two numbers is 144. What are the numbers ? 

2. The sum of certain two numbers is 135. The greater is 
14 times the smaller. What are the numbers ? 


LITERAL NUMBERS — FORMULAE 


21 


3. James’ age is 5 times Fred’s age. The sum of their ages 
is 54 years. How old is each ? 

4. A man owns two farms having a total of 225 acres. The 
second farm is 4 times as large as the first. How large is each 
farm? 

5. Two boys made a profit of $3.80 from selling lemonade. 
They had agreed that the second should receive 3 times as 
much as the first. How much should each receive ? 

6. The second of certain three numbers is 3 times the first; 
the third is 5 times the first. The sum of the three numbers 
is 126. What are the numbers ? 

7. The sum of certain three numbers is 480. The second 
is 4 times the first, and the third is 5 times the first. What are 
the numbers? 

8. The sum of certain three numbers is 260. The second is 
3 times the first; the third is 3 times the second. What are 
the numbers ? 

9. A man agreed to give a certain amount to a fund pro¬ 
vided a second man gave twice as much. A third man agreed 
to give twice as much as the second provided they made a total 
contribution of $350. How much did each have to contribute ? 

10. The length of a certain rectangle is to be 5 times its width. 
Its perimeter is to be 360 ft. How long and wide must it be ? 

11 . Find the three sides of the triangle 
pictured at the right if the perimeter is 
540 ft. 

12. A boy wants to save $21 in 6 
months. He decides to make his savings 
of the second month twice his savings of the first month; his 
savings of the third month 3 times his savings of the first 
month; his savings of the fourth month 4 times his savings of 
the first month; etc. How much must he save each month ? 





22 


ALGEBRA 



13. The perimeter of the quadrilateral ABCD is 275 inches. 
The side CD is twice as long as side AB ; side AD is 3 times as 
long; side BC equals the sum of sides 
AD and CD. How long is each ? 

14. The sum of the ages of a father, 
a mother, and son in a certain family 
is 60 years. The father is 6 times as old as his son, and the 
mother is 5 times as old. How old is each? 

16. A man had 4500 sq. ft. to devote to flower garden, vege¬ 
table garden, and a planting of small fruits. He decided to 
divide it so that the vegetable garden would be 5 times as large 
as the flower garden, and the fruit patch 3 times as large as the 
flower garden. How much space would that give him for each 
purpose ? 

PROBLEMS ABOUT ANGLES 

18. A straight line extends infinitely far in each of two 

directions. A b 

< -> 

This is indicated in the figure above by the arrow heads at the two 
ends of line AB. 

The part of a line on one side of a point, B, is called a half 

line or ray. A ray extends infinitely __ 

far in one direction. This is indicated B Cf 

by the arrow head at point C. 

19. An angle is the figure formed by two rays drawn from 

the same point. By 

Angle AOB at the right is a right angle. 

Find in the room some lines which form a 
right angle. Q 

Angles are measured by a small angle unit called a degree. 
A right angle contains 90 degrees. 

An angle smaller than a right angle is 
called an acute angle. _i_ ^ 









LITERAL NUMBERS — FORMULAE 


23 


An angle larger than a right angle (and 
less than two right angles) is called an 

obtuse angle. 

EXERCISE 14 

1. How many degrees are there in one half a right angle? 
What kind of angle is this ? 

2. How many degrees are there in -f of a right angle ? What 
kind of angle is this ? 

3. If an angle measures 30°, how much must you add to it 
to get a right angle ? 

4. Two angles whose sum is a right angle are called comple¬ 
mentary angles. Each of the angles is called the complement 
of the other. 

In the adjoining figure, angles a and b are comple¬ 
mentary. 

If angle a is 40°, how much is angle b? 

6. What is the complement of 60° ? 80° ? 

20° ? x°? 

6. Are angles of 20° and 65° complementary? 

7. If two angles measuring 3 x° and 5 x° are complementary, 
how much is their sum ? 

Form an equation expressing this fact and, from it, find these 
two angles. 

8. Form an equation expressing the fact that two angles 
measuring 2 x° and 7 x° are complementary, and then find the 
size of each of these angles. 

9. a. If an angle contains x° and its complement is 9 times 
as large, what represents its complement ? 

b. Form an equation expressing the fact that these two angles 
are complementary, and then find the angles. 

10. Find the angle which is four times as large as its comple¬ 
ment. (Let C = the number of degrees in the complement.) 







24 


ALGEBRA 


11 . Find the angle which is two 
thirds of its complement. 

12. In the figure at the right 
how many degrees are there in 

angle AOC? in angle COR? How -g- 

many degrees are there in the sum 
of these two angles ? __ 

Angle AOB is called a straight O A 

angle. 

A straight angle equals two right angles and measures 180 
degrees. 


13. Find the size of each angle in the adjoining figure. 

14. There are three angles whose sum is 180°. 

The second is 3 times the first, and the third is 
5 times the first. How large is each ? 

15. a. What size of angle must be added to an angle of 60° 
to make a straight angle ? 

b. Two angles whose sum is a straight angle are called supple¬ 
mentary angles. Each of the angles is the supplement of the 
other. 



16. What is the supplement of 40° ? of90°? of 150°? ofz°? 

17. Form the equation expressing the fact that two angles 
measuring 4 x° and 8 x° are supplementary. Solve this equation 
and then determine the size of the angles. 

18. a. Find the angle which is 8 times as large as its supple¬ 
ment. 

b. What kind of angle is each ? 

19. Find the angle which is 11 times as large as its supplement. 

20. a. If you examine Figure 1 (p. 25) you will observe that 
the sum of the four angles is 360°. In Figures 2 and 3, the sum 
of the four angles must also be 360°. 

The sum of the angles around a point is 360°. 






LITERAL NUMBERS — FORMULA 


25 


b. In Figure 3, since the sum of the angles must be 360°, form 
an equation expressing this fact, solve it, and find the size of 
each of the angles. , 


C 

A- (i -B 

D 




Fig. 1 Fig. 2 Fig. 3 

21. There are four angles making up the total angle around 
a point. The second angle is 3 times the first; the third is 
one half the first; the fourth is 3 times the third. How large 
is each of these angles ? 

22. a. Draw a large triangle on 
paper, and then cut it from the paper. 

Tear off the three corners as illus¬ 
trated at the right. Place these 
corners side by side as shown in the 
second figure at the right. You will 
find that the sum of angles A , B, and C is 180°. 

You can prove this fact by measuring the angles, if you know 
how to use a protractor. 





The sum of the angles of a triangle is 180°. 


b. If the three angles of a triangle measure x°, 2 x°, and 3 x°, 
write the equation expressing their sum, solve the equation, and 
find the size of each angle. 

23. How large is each of the angles of a triangle if the second 
is equal to the first and the third is 3 times as large as the first ? 

24. Find the three angles of a triangle if the second is one 
third of the first and the third is two thirds of the first. 

25. Find the three angles of a triangle if the second and the 
third are each one half of the first. 


26. Find the three angles of a triangle if two angles are equal 
to each other and the third angle is twice their sum. 






26 


ALGEBRA 


EVALUATING ALGEBRAIC EXPRESSIONS 

20. An algebraic expression is a number expressed by literal 
and arithmetical numbers, connected by mathematical signs 
which tell what to do with the numbers. 

Thus, 2 x — 3 yz is an expression. 

x is to be multiplied by 2; from the result, subtract 3 times the 
product of z and y. 

The arithmetical value of an expression is found by substitut¬ 
ing arithmetical values for the literal numbers, and doing what 
the signs direct. 

Thus, if x = 15, y = 2, and z = 3 in 2 x — 3 yz, 
then 2z — 3yz = 2xi5 — 3X2X3 
= 30-18 
= 12. 

21. Order of fundamental operations. Addition, subtraction, 
multiplication, and division are called the four fundamental opera¬ 
tions. Often several of them must be performed when finding 
the value of an expression, — as in the example above. 

Rule. To evaluate a complicated expression involving the 
four fundamental operations: 

1. First, do all the multiplications. 

2. Second, do all the divisions, taking them in order from 
left to right. 

3. Then do the additions and subtractions, taking them in 
any order. 

The example above was solved according to this rule. If, instead, 
we had first subtracted 3 from 15, getting 12; had multiplied this by 
2, getting 24; had multiplied this by 2 and by 3, the final result would 
have been 144. The correct result is 12. 

22. Power, base, and exponent. When the same number is 
used as a factor several times, the result is called a power of 
the number; the number itself is called the base; a small 




LITERAL NUMBERS — FORMULAE 


27 


integer written at the right and above the base telling how 
many times the base is used as a factor is called an exponent. 

Thus, x 2 means x • x. Read it “x square” or “x second power.” 
x 3 means x • x • x. Read it “x cube” or “x third power.” 

. x 4 means x • x • x • x. Read it “x fourth” or 11 x fourth power.” 
The number x is the base in each of these illustrations; the numbers 
2, 3, and 4 are exponents. 


EXERCISE 15 

Find the numerical value of the following expressions when 
a = 2, 6=4, c = 5, d = 3, x = 2, y = 5: 


1. 

a 2 

6. 

2 cd -+■ xy 

11. 

a 2 + ab + b 2 

2. 

b z 

7. 

7 x + 5 y 

12. 

x 2 + 5 x + 6 

3. 

X 4 

8. 

9 a — 4 x 

13. 

a 3 + 6 3 — ab 

4. 

xy 

9. 

a 3 -x 3 

14. 

(5 a + 3 b) -¥■ xy 

5. 

ah 2 

10. 

y 2 -b 2 

15. 

(cd — 5 a) -h bx 


Each of the expressions above can be expressed in words. 
Thus, the 13th can be stated: 

the cube of a, increased by the cube of b, diminished by the product of 
a and b. 

16-27. Express in words, preferably orally, the algebraic 
expressions given in Examples 1-12 above. 

Write an algebraic expression to represent: 

28. The sum of a and b. 

29. The product of x and y. 

30. The square of x increased by the cube of y . 

31. 5 times the product of c and d. 

32. The cube of a decreased (made smaller) by the cube of b. 

33. The square of x plus 8 times x, diminished by 7. 

34. The quotient of a divided by b. 

35. The quotient of c divided by d, minus the quotient of x 
divided by y. 


28 


ALGEBRA 


23. Making and evaluating formulae is one of the useful ap¬ 
plications of the previous section. 

A rule of computation expresses the relation of one number 
to one or more other numbers. 

Thus, the area of a trapezoid equals one 
half its altitude multiplied by the sum of 
the bases. 

This rule gives the relation of the area to 
the lengths of the altitude and the bases of the trapezoid. 

This same rule expressed in algebraic symbols is a formula. 

Let a = the number of linear units in the altitude, 

b = the number of the same units in the upper base, 
c = the number in the lower base, 

and A = the number of the corresponding surface units inside 
the trapezoid; 

then A = i a(b + c). 

This formula also expresses the relation of the number A to the 
numbers a, b, and c. Review the manner of using a formula as 
taught in § 5. 

EXERCISE 16 

1. Using the formula for the area of a trapezoid, find A 
when a = 18, b = 20, and c = 30. (See § 5.) 

2. Find the area of a trapezoid whose altitude is 8 in., whose 
upper base is 11 in., and whose lower base is 15 in. 

3. In simple interest problems, the amount equals the prin¬ 
cipal plus the interest. 

a. Express this relation as a formula, using A for the amount, 
P for the principal, and I for the interest. 

b. Using the formula you made in Step a, find A when P = 
$600 and I = $16.50. 

c. Also find A when P — $800 and I = $25. 






LITERAL NUMBERS — FORMULAE 


29 


d. If P remains the same in value, but I increases, what 
happens to A ? 

e. If I remains the same in value and P increases, what hap¬ 
pens to A ? 

4. The selling price of an article equals its cost plus the gain 
made when selling it. 

a. Express this relation between the selling price, S, the 
cost, C, and the gain, G. 

b. Using your formula, find S when C = $6.50 and G = $2.25. 

c. Also find S when C = 89^ and G = 236. 

6. You probably know that the area of a circle equals pi 
(3.1416 or 3^) times the square of the radius. 

a. Express this rule as a formula, using A for the area, x for 
the number pi, and r for the radius. 

b. Find, by your formula, the area A when r = 7. 

c. Find the area of the circle whose diameter is 16 in. 

6. It is proved in geometry that the volume, V, of a sphere 
equals four thirds of pi times the cube of the radius, r. 

a. Express this relation between V, x, and r as a formula. 

b. Using your formula, find V when r = 3. 

c. Find tjie volume of the sphere whose radius is 5. 

7. Each of n bags of grain weighs k pounds. Let w be their 
total weight. 

a. Express in a formula the relation between w, n, andk. 

b. By your formula find w when n = 25 and k = 100. 

8. F = Mv 2 /gr. (Formula from physics.) 

Find F when M = 175, v — 25, g = 32, and r = 5. 

9. F = f C + 32. (From the study of heat in physics.) 

Find F when C = 90. 

10. A = P(1 + rt). (Formula from the study of simple 
interest.) 

Find A when P = $5000, r = .06, and t = 3|. 


30 


ALGEBRA 


11. S = at + \ gt 2 . (Formula from mechanics.) 

Find S when a = 500, t = 3, and g = 32. 

12. C = nE/(R + nr). (From the study of electricity.) 
Find C when E = 1.08, n = 40, R = 500, and r = 4. 

13. d = z‘/(l + i). (From the mathematics of investment.) 
Find d when i = .06. 

14 . S = P( 1 + r) 2 . (From the study of compound interest.) 
Find S when P = $2500 and r = .05. 

15. p = ad?/t + d. (From an engineer’s manual.) 

Find p when a = .56, d = and t = f. 

Note. — Before starting Chapter II, each pupil should memorize 
the formulae collected in § 6, p. 8, and those given in Exercise 16, 
Examples 1, 3, 5, and 6. 

I.e. Example 1. A = \ a(b + c) 

Example 3. A = P + I 
Example 5. A = irr 2 
Example 6. V = $ -rrr 3 

Supplementary problems on formulae appear on page 426. , 


n. POSITIVE AND NEGATIVE NUMBERS 


24. Another characteristic of algebra is the use in it of 
numbers called positive and negative numbers. 

These numbers are used to distinguish between opposite 
quantities. 

Thus, rising and falling temperature are opposite quantities. It is 
not enough to be told that the temperature changed 5 degrees (5°) in a 
certain time; it is necessary to know whether the temperature rose 5° 
or fell 5°. 

A 5° rise in temperature followed by a 5° fall has a total effect of no 
change at all. 

Similarly, a $50 gain and a $50 loss are opposite quantities. A gain 
of $50 followed by a loss of $20 has the same effect as a single gain of 
$30. 


EXERCISE 17 

What is the opposite of: 

1. Sailing 25 miles north ? 

2. Walking 10 steps forward? 

3. Gaining 15 pounds in weight? 

4. Depositing $20 in a bank ? 

6. A 10° fall in temperature ? 

6. Adding 50? 

What is the total effect of: 

7. First winning 15 points in a game and then losing 10 
points ? 

8. First walking 12 steps forward and then 8 steps back¬ 
ward? 

9. A 10° rise in temperature followed by a 15° fall ? 

31 


32 


ALGEBRA 


10. Walking 20 steps to the right and then 25 steps to the 
left? 

11. An 8° rise in temperature followed by a 15° fall? 

12. Driving an automobile 25 mi. north and then 30 mi. 
south ? 

13. First adding 26 and then subtracting 30? 

14. First adding 18 and then subtracting 10 ? 

16. First adding 18 and then subtracting 25? 

25. The signs -f and — are used to distinguish between 
opposite quantities. 

Thus, 8° above zero is written + 8°; and 8° below zero is written 
- 8 °. 

Similarly, a $5 gain may be written + $5; and then a $5 loss, — $5. 

A quantity preceded by a plus sign is called a positive quan¬ 
tity ; one preceded by a minus sign is called a negative quantity. 


EXERCISE 18 


1. The temperature at 8 a.m. was + 10°. At 
noon, it was 15° warmer. What was the tempera¬ 
ture then? 

2. a. At 7 a.m., the temperature was +25°. The 
temperature rose 8° by 10 a.m. What was the tem¬ 
perature then? 

b. By 6 P.M., the temperature was 15° lower than 
it was at 10 a.m. What was the temperature at 
6 p.m. ? 

3. a. At 7 a.m., the temperature was — 5°. By 
noon it had risen 5°. What was the temperature 
then? 

b. By 6 p.m., it had fallen 8° from the temperature 
at noon. What was it at 6 p.m. ? 















POSITIVE AND NEGATIVE NUMBERS 33 

4. Let us agree to indicate walking to the right by a plus 
sign, and walking to the left by a minus sign. 

What is the total effect of + 10 steps followed by — 15 steps ? 

Let us agree to write the solution as follows : 

(+10 steps) + (— 15 steps) = — .5 steps, 
but read it as follows: 

10 steps to the right, followed by 15 steps to the left, give the same 
effect as 5 steps to the left. 

Similarly, read and tell the result of: 

5. (+15 steps) + (— 10 steps) 

6. (+15 steps) + (— 20 steps) 

7. (+15 steps) + (— 15 steps) 

8. (—5 steps) + ( + 8 steps) 

9. (— 8 steps) + (— 2 steps) 

10. a. If a positive sign indicates a gain , what does a negative 
sign indicate ? 

b. What is the total result of 

(+ $15) + (+ $20) + ( — $25)? 

Note. — Read this, “a $15 gain, followed by a $20 gain, followed 
by a $25 loss.” 

Similarly, read and give the result of: 

11. (+ $20) + (+ $10) + (— $15) 

12. (+ $50) + (- $40) + (+ $10) 

13. (+ $100)* + (- $25) + (- $40) 

14. (- $50) + (- $10) + (- $15) 

15. An automobile was driven + 15 mi. (15 mi. north) from 
a starting point and then — 8 mi. What was its final distance 
from the starting point ? 

16. What was the final distance from its starting point of a 
taxi which traveled + 3 mi., + 7 mi., — 15 mi., + 4 mi., 
+ 16 mi. ? 


34 


ALGEBRA 


17. If a boy marks money he receives by a plus sign, and 
money he spends by a minus sign, how much should he have 
on hand to agree with the following record: 

+ $1.35, + 256 -5ty, +406 +356 - 65^? 

18. A grain dealer marked by a plus sign grain he purchased, 
and, by a minus sign, grain he sold. How much should he have 
on hand to agree with the following record: 

+ 50 bu., + 75 bu., — 10 bu., — 5 bu., + 25 bu., — 8 bu. ? 

19. A teacher marked with a plus sign the number of pupils 
who enrolled in her room, and by a minus sign the number who 
dropped out. How many should be left after she made the 
following record: 

(+ 35) + (+ 2) + (- 1) + (- 3) + (+ 1)? 

26. Positive and negative numbers. In algebra, instead of 
an arithmetical number, like 3 , there is the number + 3 , called 
positive 3 , and the number — 3 , called negative 3 . 

The fundamental fact about the numbers + 3 and — 3 is that 
(+3) + (-3) = 0. 

Similarly, (+ f) + (- f) = 0; also (+ x) + (— x) = 0. 

Since ( — 3) + (+ 3) =0, it is natural to think of — 3 as 
being less than zero. 

The following geometrical representation of positive and 
negative numbers aids in understanding them. 

—5 -4 -3 -2 -I 0 +1 -4-2 -4-3 -4-4 +5 


0 is represented on this line by point A. 

+ 1, +2, +3, etc., are represented by points to the right. 

— 1, — 2, — 3, etc., are represented by points to the left of the zero 
point. 

Notice that any positive number precedes all larger positive num¬ 
bers. Since all points marked by negative numbers precede those 
marked by zero or the positive numbers, every negative number is less 
than zero and less than any positive number. 

Thus, — 5 is less than + 1. 



POSITIVE AND NEGATIVE NUMBERS 


35 


The sign must always be written before a negative number; 
if no sign appears before a number, the number is considered 
positive. 

The arithmetical number which remains when the sign of a 
positive or of a negative number is omitted is called the absolute 
value of the number. 

Thus, the absolute value of + 5, or of — 5, is 5. 

Historical Note. — Hindu mathematicians, who knew about posi¬ 
tive and negative numbers long before European mathematicians, in 
referring to them, used words which correspond to our words for assets 
and debits. They also were acquainted with the illustration of such 
numbers by means of the opposite directions on a straight line. To 
indicate that a number was a negative number they placed a dot over 
it. European mathematicians did not arrive at an equal understanding 
of positive and negative numbers until the sixteenth century. 

EXERCISE 19 

Read the following numbers and tell the absolute value of 
each. 

1. + 15 3. + f 5. + -6 7. + 2£ 9. - T \ 

2. - 60 4. -f 6. - .08 8. - 3.07 10. + .075 

11. a. Which is the largest of the numbers in Examples 1-10 ? 

b. Which has the largest absolute value ? 

12. a. Which of the numbers in Examples 1-10 is the small¬ 
est? 

b. Which has the smallest absolute value ? 

27. Addition of positive and negative numbers. 

Example 1. — Find the sum of + 6 and + 4. 

Solution. — Just as $6 gain followed by $4 gain gives $10 gain, so 
(+ 6) + (+ 4) = + 10. 

Example 2. — Find the sum of — 6 and — 4. 

Solution. — Just as $6 loss followed by $4 loss gives $10 loss, so 
(-6) + (-4) = -10. 


36 


ALGEBRA 


Example 3. — Find the sum of + 6 and — 4. 

Solution. — Just as $6 gain followed by $4 loss gives $2 gain, so 
(+6) + (-4) =+2. 

Example 4. — Find the sum of — 6 and 4- 4. 

Solution. — Just as $6 loss followed by $4 gain gives $2 loss, so 
(-6) +(4-4)= -2. 

Rule. — 1. To add two positive numbers, add their absolute 
value ( § 26) and prefix the plus sign to the result. (Example 1.) 

2. To add two negative numbers, add their absolute values 
and prefix the minus sign to the result. (Example 2.) 

3. To add a positive and a negative number, 

Subtract the smaller absolute value from the larger and prefix 
to the result the sign of the number having the larger absolute 
value. (Examples 3 and 4.) 

EXERCISER) 

1. a. Add 4- 3 to each of the following numbers: 

4-8, 4-7, 4- 12, 4“ 16, + 20, 4- 32. 

b. Give the rule which you use in this example. 

c. What is the sign of the sum when you add positive num¬ 
bers? 

2. a. Add — 5 to each of the following numbers: 

- 10, - 7, - 8, - 2, - 20, - 26. 

b. Give the rule which you use in this example. 

c. What is the sign of the sum when you add two negative 
numbers ? 

3. a. Add — 15 to each of the numbers in Example 1. 
b. Give the rule which you use in this example. 

4. Add -f 16 to each of the numbers in Example 2. 

5. Add 4- 12 to each of the following numbers : 

4- 15, - 5, 4- 11, - 3, - 9, 4-7. 


POSITIVE AND NEGATIVE NUMBERS 


37 


6 . Add — 9 to each of the numbers in Example 5. 

Note. — For further drill, add other numbers to the numbers in 
Example 5 ( e.g . + 16, — 5, etc.). 


Find the following sums : 


7. 

8. 

9. 

10. 

11. 

12. 

+ 92 

- 74 

-85 

+ 106 

- 240 

- 10.5 

- 36 

+ 60 

- 37 

- 72 

+ 155 

- 6.4 

13. 

14. 

15. 

16. 

17. 

18. 

- 6 

+ 15 

- 24 

+ 68 

- 84 

+ 9.5 

+ 22 

- 35 

+ 40 

- 96 

+ 107 

- 15.6 

19. 

Find + 5 

- 4 

• +7 

- 9 

Hint. — Either add in order from the bottom 
to the top; or, first add the positive numbers, 
obtaining + 12 ; then, add the negative num¬ 
bers, obtaining — 13; then add these two 
results. 

20. 

21. 

22. 

23. 

24. 

25. 

+ 5 

- 6 

+ 12 

- 6 

+ 14 

- 8 

- 7 

- 4 

- 10 

- 11 

- 12 

+ 11 

- 3 

+ 8 

+ 5 

+ 10 

- 9 

- 12 

+ 4 

- 2 

+ 7 

+ 5 

+ 20 

+ 9 

- 5 

+ 7 

- 8 

- 7 

- 7 

- 4 


Find the sum of : 

26. — \ and + -J 28. — f and — \ 30. + 2 \ and — 3^ 

27. + f and — 29. + f and — yu 31. — and + 2f 

28. Adding a negative number gives the same result as sub¬ 
tracting the corresponding arithmetical number. 

Thus, 12 + ( — 3) gives the same result as 12 — 3. Each equals + 9. 

29. Multiplication of positive and negative numbers. 

The words multiplier, multiplicand, and product have the 
same meaning in algebra as in arithmetic. 


















38 


ALGEBRA 


Rules for multiplication of signed numbers are suggested by 

the following examples. 

The sign X is to be read “ multiplied by” 

Example 1. — Find ( + 4) multiplied by (+ 3). 

Solution. — (+ 4) X ( + 3) = (+ 4) + (+*4) + ( + 4) = + 12. 

/. (+ 4) X (+ 3) = + 12. 

Example 2. — Find (—4) multiplied by ( + 3). 

Solution. — (- 4) X (+ 3) = (- 4) + (- 4) + (- 4) = - 12. 

.*. (- 4) X (+ 3) = - 12. 

Example 3. —Find (+4) multiplied by (— 3). 

Solution. — 1. In arithmetic, 4 X 3 = 3 X 4; that is, the multiplier 
and multiplicand may be interchanged. 

2. If it is assumed that the same law holds for negative numbers, then 

(+4) X (—3) should equal (—3) X ( + 4). 

3. But (- 3) X (+ 4) = (- 3) + (- 3) + (- 3) + (- 3) = - 12. 

4. /. also ( + 4) X (- 3) = - 12. 

Note. — When multiplying by a negative number, — 3, it appears 
that the result, — 12, may be obtained by multiplying first as if the 
multiplier were positive, + 3, getting + 12, and then changing the 
+ 12 to its opposite, — 12. 

Example 4. —Find (—4) multiplied by (— 3). 

Solution. — 1. Using the idea presented in the note following Ex¬ 
ample 3, first multiply (—4) by (+3). This gives —12. (See 
Example 2.) 

2. Now take the opposite of — 12. This gives + 12. 

3. (-4) X (-3) = +12. 

Rule. — To multiply one signed number by another: 

1. Find the product of their absolute values. (See all four 
examples above.) 

2. Make the product positive if the multiplicand and multiplier 
have like signs. (See Examples 1 and 4 above.) 


POSITIVE AND NEGATIVE NUMBERS 


39 


3. Make the product negative if the multiplicand and multi¬ 
plier have unlike signs. (See Examples 2 and 3 above.) 


EXERCISE 21 


1. Multiply the following numbers by + 5: 

+ 3, +7, +9, +6, +11, + f, + £ 

2. Multiply the numbers in Example 1 by — 3. 

3. Multiply the following numbers by + 4: 

- 7, - 10, - 5, - 8 - 12, - f, - £ 

4. Multiply the numbers in Example 3 by — 5. 

6. Multiply the numbers in Example 1 by — 6. 

6. Multiply the numbers in Example 3 by + 10. 

7. Multiply the numbers in Example 1 by — 4. 

8. Multiply the numbers in Example 3 by — 6. 

9. What is the sign of the product when the multiplier and 
multiplicand have the same sign? when they have unlike 
signs ? 

10. Multiply each of the following numbers by — 8: 

+ 6, - 10, + 12, - 9, - 4, - b +f 

11 . Multiply the numbers in Example 10 by + 12. 

12. Multiply the numbers in Example 10 by + 16. 

13. Multiply the numbers in Example 10 by — 20. 

14. Multiply the following numbers by — 24: 


1 il l 5 _ 2 _ 7 

— 2> ~i -3 j — 4> — ‘3’ TH¬ 


IS. Multiply the numbers in Example 14 by + 36. 


Find the following products : 


16. (+ .7) X (- 1.5) 21. (- 3) X (+ 2) X (- 4) 

17. (- 6) X (+ .25) 22. (+ 5) X (- 5) X (+ 5) 

18. (- .4) X (- .8) 23. (- 6) X (- 2) X (- 3) 

19. (+ 2.5) X (- 1.5) 24. (- 1) X (+ 4) X (- 2) 

20. (- 3.2) X (- 1.2) 25. (+ 7) X (- 5) X (+ 2) 




40 


ALGEBRA 


30. Powers of positive and negative numbers. 

(+ 2) 2 = (+2) -.(+2) = + 4 
(+ 2) 3 = (+ 2 )-(+ 2 )-(+ 2 ) = + 8 
(+ 2 y = (+ 2) • (+ 2) • (+ 2) • (+ 2) = + 16 

Clearly, every power of a positive number is positive. 

(-2) 2 = (-2)-(-2) = +4 
(-2) 3 = (-2)-(-2)-(-2) = - 8 
(- 2) 4 = (- 2 ) • (- 2 ) . (- 2 ) • (- 2 ) = + 16 
(- 2) 5 ='(- 2) • (- 2) • (- 2) • (- 2) • (- 2) = - 32 

31. Clearly, every even power of a negative number is positive, 
and every odd power is negative. 


EXERCISE 22 

Find the values of the following: 

1. (— 3) 2 4. (— 3) 5 7. (— 2) 6 10. (+ 1) 6 

2. (— 3) 4 5. (— 4) 2 8. (— 5) 3 11. (-1) 5 

3. (— 3) 3 6. ( — 4) 3 9. (— 5) 4 12. (- l) 6 

13. Find the value of ax 2 + bx + c 

when a = 5, b = 2, c = — 3, and x = — 2. 

Solution. — l.ax 2 + bx + c=5 •(- 2) 2 + 2(- 2) + (- 3) 

2. =5-(+4) + (-4) + (-3) 

3. =20-4-3 

4. = 13. 

Note. — Observe that +( — 4) gives — 4; and +( — 3) gives — 3, 
according to § 28, page 37. 


Find the values of the following expressions, when a = 3, 
b = — 2, c — — 3, and x = — 2. (Read again § 21, page 26.) 

14. 5 ac 18. 2 x 2 + 3 x 22. 3 x 2 -f 2 x -f 1 

15. a?b 19. ax 2 — bx 23. ax 2 + bx -f- c 

16. be 2 20. bx 2 + cx 2 24. x 2 + 2 ax + a 2 

17. 3 bx 21. ax? — bx? 25. b 2 — 2 be c? 

Note. —Attention is called to the tests on pages 453-454. 


III. ADDITION AND SUBTRACTION OF 
POLYNOMIALS 


32. *A monomial or term consists of numbers connected only 
by signs of multiplication or division. Thus, 

2 x 2 , — 3 ab, and + 5 are the terms of the expression 2 x 2 — 3 ab + 5. 

If the sign before a term is plus, as 2 x 2 , it is called a positive term; 
if minus, as — 3 ab, it is called a negative term. 

33. If two or more numbers are multiplied together, each of 
them or the product of any number of them is a factor of the 
product. 

Thus, a, b, c, ab, ac, and be are all factors of abc. 

34. Any factor of a product is called the coefficient of the 
product of the remaining factors. 

Thus, in the term 2 ab, 2 is the coefficient of ab, 2 a of 6, and 2 6 of a. 

35. In a term like 5 x 2 y, the factor 5 is called the numerical 
coefficient of x 2 y. 

A term like x has the numerical coefficient 1, since x is the same as 1 x. 

If a term is negative, like —3 a, the numerical coefficient is — 3, 
since (— 3) • a = —3 a. 

36. Like terms are terms which have the same literal factors. 

Thus, 2 x 2 y and — 5 x 2 y are like terms; also, 2 and 3 • 

Unlike terms are terms which do not have the same literal 
factors. 

Thus, 2 x 2 y and 2 xy 2 are unlike terms. 

Unlike terms may be like with respect to one or more letters. 

Thus, 2 axy and 3 byx are like with respect to xy. 

41 


42 


ALGEBRA 


EXERCISE 23 

1. Give five different factors of 8 xy. 

2. What is the coefficient of xy in 8 xy ? of y ? 

3. What is the numerical coefficient of ab in — 6 ab? 

4. Are 8 x 3 and — 7 x z like terms ? 

6. Are 8 x 2 and — 7 x? like terms ? 

37. Addition of like terms with positive coefficients has been 
taught previously. 

Thus, 6 n + 4 n = 10 n. 

n is the common factor of 6 n and 4 n. 6 and 4 are the coefficients 
of n. Their sum, 10, is the coefficient of the result. 

Rule. To add two or more like terms, multiply their common 
factor by the sum of its coefficients. 

This rule is employed also when the coefficients are negative. 

Example. — What is the sum of — 5 x 2 y and + 3 x 2 y ? 

Solution. — 1. The coefficients of x 2 y are - 5 and + 3, and the sum 
of these coefficients is — 2. 

2 - •*• (—5 x 2 y ) -f (+ 3 j?y) = — 2 x 2 y. 

No matter what the values of x and y may be, the sum of 
(—5 x 2 y) and ( + 3 x 2 y) is always — 2 x 2 y. 

Thus, if x = 2 and y = 1 

- 5 x 2 y = -5-41= - 20; 3 x 2 y = 3 • 4 • 1 = 12; 

and (- 20) + (+ 12) = - 8. 

Also, — 2' x 2 • y = —2-4-1= — 8. 

For this reason, the correctness of addition of polynomials 
can be checked by substituting numbers. 


EXERCISE 24 


Find the sum of: 

1. 8 A and — 15 A 

2. 9 p and — 12 p 

3. —11m and -f 8 m 

4. — 15 t and — 4 t 


5. 16 E and -27 E 

6. — 17 r^s and + 26 r 2 s 

7. - xyz and + 14 xyz 

8. - be 2 and - 7 6c 2 




ADDITION AND SUBTRACTION OF POLYNOMIALS 43 


Time yourself to see how quickly you can get the sums in the 
following examples: 


9. 

10. 

11. 

12. 

13. 

- 18 p 

+ 17 x 2 y 

— 5 rs 

+ 9w> 2 

— 28 xyz 

+ 11 p 

— 4 x 2 y 

— 16 rs 

- 36 jo 2 

+ 42 xyz 

14. 

15. 

16. 

17. 

18. 

- 5 1 

+ 17 x 2 

+ 9 ab 

+ xyz 

7 ab 2 c 

+ 12 t 

CO 

1 

- 10 ab 

— 4 xyz 

— ab 2 c 

- 9 t 

+ 8x 2 

- 13 ab 

+ 6 xyz 

— 8 ab 2 c 

- 4 t 

- 12 x 2 

+ 7 ab 

— 13 xyz 

+ 14 ab 2 c 

+ 6 1 

+ 5x 2 

+ 5 ab 

+ 5 xyz 

— 9 ab 2 c 

19. 

20. 

21. 

22. 

23. 

— ™ 

-%rs 

- t ** 

- 2.5 ab 

- .25 y 2 

+ J mn 

+ %rs 

+ i 

+ 3.4 ab 

-3.5 i/ 2 

— mn 

-irs 

_ 4 

TO x 

— .6 ab 

+ 4.5 y 2 


38. Adding unlike terms. If asked to add 5 gal. and 3 qt., 
you can write 5 gal. + 3 qt. This expression indicates the sum. 
You get the actual sum only when you change 5 gal. to 20 qt. 
Then, 5 gal. + 3 qt. = 20 qt. + 3 qt. = 23 qt. 

Similarly, if asked to add 5 x and 3 y , you can only indicate 
the sum by writing 5 x + 3 y. 

You cannot combine 5 x and 3 y in any way since they are 
unlike terms. 

Later, if you learn that x = 4 and y = 6, you can find the 
value of the sum. Then 5 x + 3 y = 20 + 18 = 38. 

Addition of unlike terms can only be indicated, and that is 
done by writing them with a plus sign between them. 

39. Recall that adding a negative number has the same effect 
as subtracting the corresponding positive number. 

Thus, 10+(-4) = 10-4; 

and 5x+(-2x) = 5x-2x. 

















44 


ALGEBRA 


Example. — a. Write in simplest form the sum of 2 a; 2 , — 3 x, 
— 4, — 5 x 2 , + 7, and + 8 x. 

Solution. — 1. Indicating the addition, the sum is 

( 2x 2 ) + (_ 3 x) + (- 4) + (- 5x 2 ) + (+ 7) + (+8A 

2. And this is the same as, 

2 x 2 - 3 x — 4 - 5 x 2 + 7 + 8 x. (§28) 

3. Combining like terms, the sum is — 3 x 2 + 5 x + 3. 

6. What is the value of the sum, when x = 2 ? 

Solution. — 1. —3x 2 + 5x + 3= (-3-4) + 5- 2+ 3 
2. = - 12 + 10 + 3 = + 1. 


EXERCISE 25 

Indicate the sum of the terms in each of the following ex¬ 
amples, and simplify the result by combining like terms: 

1. 7 ah, - 3 ab , + 4 ab 6. 16 t 2 , - 11 t 2 , - 12 t 2 

2. 9 r 2 , - 7 r 2 , + 4 r 2 , - 8 r 2 6. — \ xy, £ xy, - § xy 


3. 12 xy, — 10 xy, + 6 xy 

4. — 4 z, + 7 z, — 10 s 


3 ' m , ~ Q m > ~ T 2 m 
- f W,\w, + A w 


9. a. 8 a; 2 , — 9 a;, — 6 a; 2 , + 5, + 4 a;, — 7 
b. What is the value of the sum, when x = 2 ? 

10. a. 4 o, - 5 6, - 3 c, + 7 ft, + c, - 2 a 

6. What is the value of the sum, when a = 3, 6=2, c = l? 

11. 10 m 3 4 , + 6 n 2 , — 8 mn, — 4 m 3 , + 5 ran, and — 2 n 2 

12. 10 r 2 , — 6 r, + 5 5, - 8 r, + 2 s, — 5 r 2 , and — 4 5 

13. 9 r 3 , - 6 r 3 , + 3 r 2 , - 7 r, + 8 r 2 , - 15>, and + 5 

14. 2.5 xy, — 1.2 xy, + .5 xy, and — 3.8 xy 

15. 2.4 a 3 , - 5 a 2 , + .7 a 3 , - 3, + 2 a 2 , and - 1 


40. A polynomial is an algebraic expression of two or more 
terms, as, a + 6 or x 2 — 2 xy + 5 y 2 . 

A binomial is a polynomial having two terms; as a + 6. 

A trinomial is a polynomial having three terms; as x 2 — 2 xy 
+ 5y 2 . 


ADDITION AND SUBTRACTION OF POLYNOMIALS 45 


Note. — If a number expression is inclosed in parentheses, it is 
considered a term. 

Thus, a + {b + c) is, for the time, a binomial having the term a and 
the term (b + c). 

41. Arrangement of terms in ascending and descending 
powers of a letter. 

If the terms of a polynomial are — 3 x 2 y, + x 3 , — y s , and 
+ 3 xy 2 , a natural orderly arrangement of them is as follows : 
x 3 — 3 x 2 y + 3 xy 2 — y z . 

These terms are arranged in descending powers of x. 

The exponents of x in order are now 3, 2, 1, and 0, because 
— y z does not have any factor x. 

The terms are arranged in ascending powers of y. 

The exponents of y in order are 0, 1, 2, 3, for the first term, 
x 3 , does not contain any factor y. 


42. Addition of polynomials is like addition of denominate 
numbers in arithmetic. 

Example 1. — Add 3 yd. 2 ft. 6 in. and 5 yd. 4 in. 

Solution. 3 yd. 2 ft. 6 in. 

5 yd._4 in. 

8 yd. 2 ft. 10 in. 

In this solution, like units are placed in the same column. 
Example 2. — Add 4 a + 5 6 —2c and —3 a 2 b — 6 c. 

Solution. 4 a + 5 5 — 2c 

— 3 ci + 2 5 — 6c 

a + 7 5 — 8c 


In this example, a , b, and c may represent any numbers. 

Thus, if a = 2, b = 3, and c = 1, 

4 a + 5 — 2c = 8 + 15 —2 = 21 

— 3 a + 2 6 — 6c = — 6 + 6— 6 
Also, the sum a + 7 b — 8c = 2+21 —8 = 15. 

That is, the value of4a + 55-2c, added to the value of - 3 a + 
2 b — 6 c, is the same as the value of the sum, a + 7 b — 8 c. 


2 2 \] 21 +(-6) = 15. 




46 


ALGEBRA 


This fact is most important. It can be used when checking 
a solution. 

Rule. — To add polynomials: 

1. Rewrite the polynomials, if necessary, in descending 
powers, or ascending powers of one letter, with like terms in 
the same column. 

2. Add the columns of like terms and combine the Results. 

3. Check the solution by adding again in the opposite direc¬ 
tion, or by substituting values for the letters. 

Example 3. — Find the sum of x 3 — y 3 + 3 xy 2 — 3 x 2 y, 
2 x? + 3 xy 2 + y 3 , and 2 x 2 y — 4 xy 2 . Check, by letting x = 2 
and y — 1. 

Solution. — 1. Arrange the terms in descending powers of x , with 
like terms in the same column. 

z 3 - 3 xhj + 3 xy 2 - y* = 8 - 12 + 0 - 1 = 1 

2 z 3 + 3 xy 2 -f y 3 = 16 + 6 + 1 = 23 

4- 2 x 2 y — 4 xy 2 _= 8—8 = 0 

3 z 3 — x 2 y + 2 xy 2 24 

2. Also, 3 z 3 — a*y + 2 xy 2 = 24 — 4 + 4 = 24. 

3. .*. The solution is correct. 

EXERCISE 26 


Find the following sums, checking by substitution: 


1. 

+ 7 3? - 2x + 1 
-2x 2 + 4x -3 
- 3x? + x + 2 

2. 

— 10 A + 5 B 
+44-3S 
+ 2 A -6 B 

3. 

+ 3 m — 4n +7 p 
— 5 m -f 9 n — 6 p 
+ 7m - 6 ?i — 5 p 

4. 

5. 

6. 

a 2 — 3 a + 1 

x 2 + y 2 

m 2 + 2 mn -f n 2 

a 2 + 9 a — 6 

2 x 2 + xy 

2 m 2 -3 n 2 

2 a 2 +3 

- 3 xy - 2 y 2 

— 3 mn -f 2 n 2 


Note. — Observe that there are blank spaces in Examples 4, 5, and 
6 for powers which are missing from the polynomials. 









ADDITION AND SUBTRACTION OF POLYNOMIALS 47 

Arrange the following examples as in Examples 1-6 and find 
the sums. Check each by adding in the opposite order. Time 
yourself to see how quickly you can get all of them. 

7. 6a — 56 + 7c and 4 a + 6 6 —4c 

8. 5k — 7 l + 3m, 4 /c + 3 Z — 5 m, and 8 k — 2 l 

9. 3a + 96—6 c, — 46+3c—6d, and 7c — 3 a + 4 d 

10. z 2 — 2 xy — y 2 , x 2 + 2 xy — y 2 , and 2 x 2 -f 2 y 2 — 5 xy 

11. 5 m 2 — 3 mn + n 2 , m 2 — 4 mn — 4 n 2 , and — 3 m 2 + 6 n 2 

12. a 3 + 3 a 2 b + 3 ab 2 + 6 3 and a 3 — 3 a 2 b — 6 3 + 3 ab 2 

13. 3 x 2 — 8 xy + 4 ?/ 2 , 2 xy — 4 z 2 — 7 i/ 2 , and 6 ?/ 2 + 3 xy 
+ 5 x 2 

14. 15 a 3 - 7 - 3 a + 16 a 2 and + 4 + 9 a - 20 a 2 - 10 a 3 

15. — 4 z + 2 x 3 — 5, 6 x 2 — 3 x + 6 x 3 , and 13 — 2 x s + 

9 a; — 8 x 2 

16. 2 a 2 — 3 b 2 , — 3 ab + 8 b 2 — 5 a 2 , and 9 a 2 + 6 ab — 10 b 2 

17. !a-£& + icand-ia+i&-ro c 

18. |m + fn-r, - + and m + f r - \ n 

19. 2.5 A - 6.25 B + 4.8 C and — 1.8 A + 5 B - 3.4 C 

20. 2.25 x 2 - 3.4 xy - 2 y 2 and - 1.75 x 2 + 2.1 a:?/ - 1.5 y 2 

Note. — Supplementary examples on addition of polynomials are 
to be found on page 427. 

43. Addition used in solving equations. 

In the adjoining figure, suppose the sugar S just balances the weight 
IF. If a 5 lb. weight is added to the right scale pan and 5 lb. of sugar 
to the left pan, then the two.pans will still balance. 

Also the two pans 'will not balance if a 5 lb. 
weight is added to the right pan and 3 lb. or any¬ 
thing other than 5 lb. to the left. 

Similarly, if 3 x = 60, then 3x + 5 = 60 + 5. 

Rule. — The same number can be added to both sides of an 
equation without destroying the equality. 










48 


ALGEBRA 


Example 1 . — Solve the equation x — 5 = 13. . 

Solution. — 1 . x — 5 is 5 less than x. If 5 is added to x — 5, the 
result is x — 5 + 5 or x. 

But, if 5 is added to the left side of the equation, it must also be 
added to the right side in order to keep the two sides equal. 

Therefore, add 5 to both sides of the equation. 

2. Then, x — 5+5 = 13+5. (Rule, page 47) 

3. Combining like terms, x = 18. 

Check. — Substituting 18 for x in the original equation, 

does 18-5=13? Yes. 

Then, 18 is the correct root of the equation. 

Note. — For meaning of “root of the equation,” see § 10, page 10. 

Hereafter, instead of writing in full 
“ add 5 to both sides of the previous equation ” we shall write A 5, 
and instead of “ combining like terms ” we shall write C. T. 


Example 2 . — Solve the equation 5 y — 13 = 22 . 

Solution. — 1 . 5 y — 13 = 22. 5 y _ 13 + 13 = 5 y 

2 - A 13 5 y = 35. 22 + 13 = 35 

3- D 5 y = 7 . (This should be 

done mentally.) 

Check. — Substitute 7 for y in the original equation. 

Does 5 • 7 — 13 = 22? 

Does 35 - 13 = 22 ? 

Yes. Therefore 7 is the correct root of the equation. 


EXERCISE 27 

Solve and check the following equations: 


1 . 

x — 

10 

= 5 

6. 

m — 

14 = - 

-8 

2. 

y - 

12 

= 16 

7. 

A - 

16 = ■ 

- 3 

3. 

2z 

- 5 

= 13 

8 . 

2 B - 

-21 = 

- 5 

4. 

3 t 

- 7 

= 14 

9. 

5 t — 

28 = 

-8 

5. 

4 n 

- 9 

= 27 

10 . 

6 r - 

• 40 = 

- 2 



ADDITION AND SUBTRACTION OF POLYNOMIALS 49 


11 . Solve the equation 7 x = 55 — 4 x. 

Solution. — 1 . 7 x = 55 — 4 x. 

We must remove 4 x from the right side. I 7 x -f 4 x = 11 x 

2. A 42 11 x = 55. I 55 — 4 x + 4 x = 55 

(Complete the solution and check it.) 

Note. — What does A^ in Step 2 mean? 


12. 

5 x = 

00 

to 

1 

CO 

15. 

0 

CO 

r-H 

II 

§ 

00 

— 5 w 

13. 

II 

t^ 

27-2 t 

16. 

CO 

i-H 

II 

CO 

r-H 

-10 A 

14. 

3 y = 

to 

1 

17. 

25 v = 8 - 

15 v 

18. 

Solve 

the equation 3 

x — 42 = 

— 4 x. 


Solution. — 

II 

<N 

1 

CO 

- 4 x. 

I 3 x - 42 + 42 = 


We must remove — 42 from the left side. I — 4x-H 42 = 42 — 4x 
2 . A 42 3 x = 42 - 4 x. 


(Complete the solution.) 


19. 

5 y - 

- 70 = 

-2 y 

25. 

3 C 

1 

CO 

r-H 

II 

CO 

1 

3 

20. 

7 m 

- 80 = 

— 3m 

26. 

4 d 

- 12 = 

- 3 

-5 d 

21. 

12 5 

- 95 = 

-75 

27. 

- 16 +2x 

= 9 

- 3x 

22. 

18 x 

- 24 = 

— 6 x 

28. 

7 w 

- 18 = 

6 - 

w 

23. 

9 2 - 

- 4 = - 

3)2 

29. 

13 5 

- 11 = 

4 - 

2s 

24. 

20 r 

- 24 = 

- 12 r 

30. 

5 y 

- 8 = - 

-5 y 

-3 

SUBTRACTION OF SIGNED 

NUMBERS 

AND EXPRESSIONS 

44. 

Meaning of subtraction. 







To subtract 4 from 7 means to determine the number which 
must be added to 4 to give 7. 

To subtract a from b means to determine the number which 
must be added to a to give b. 

This meaning of subtraction is used in “making change.” 

The number subtracted is called the subtrahend. 

The number from which the subtrahend is subtracted is called 
the minuend. 

The result is called the difference or remainder. 


50 


ALGEBRA 


EXERCISE 28 

What number must be added to the first number of each of 
the following examples to get the second ? 

1. 3, 10 4. 3 a, 5 a 7 . 14 c, 26 c 

2 . 8, 15 6 . 4 m,8m 8 . 2 b, 23 b 

3. 10, 19 6 . 9 t, 17 t 9. 5 x, 19 x 


How much must the temperature 

rise to change 


10 . 

+ 4° into + 9°? 

12. 

— 3° into + 

5°? 

11 . 

- 6 ° into - 2 °? 

13. 

— 5° into + 

8°? 

Find the number to be added 




14. 

(+4) + ? = + 10 

18. 

(- 10 ) + ? 

= +2 

15. 

(-5) + ? = 0 

19. 

(- 6 ) + ? = 

-2 

16. 

(-4) + ? = + 2 

20. 

(- 7) + ? = 

+ 8 

17. 

(- 6 ) + ? = + 3 

21. 

(- 15).+ r. 

= - 10 

How much must the temperature fall to change: 


22. 

+ 10 ° to + 2 °? 

24. 

+ 10° to - 5 

.°? 

23. 

+ 10 ° to 0 °? 

25. 

- 4° to - 12 

:°? 

Find the number to be added: 




26. 

(+ 5) + ? = + 3 

36. 

(- 6 ) + ? = 

+ 5 

27. 

(+ 5) + ? = 0 

37. 

(+5) + ? = 

-4 

28. 

(+ 5) + ? = - 3 

38. 

(- 7) + ? = 

- 16 

29. 

(+ 10 ) + ? = 0 

39. 

(- 10 ) + ? = 

= -4 

30. 

(+ 10) + ? = - 5 

40. 

(+ 3) + ? = 

- 11 

31. 

(- 2 ) + ?- 10 

41. 

(+7) + ? = 

- 1 

32. 

(-4) + ? = - 15 

42. 

(- 9) + ? = 

-20 

33. 

(+ 3) + ?- 6 

43. 

(- 6 ) + ? = 

+ 13 

34. 

(+ 10) + ? = - 4 

44. 

(+2)+ ? = 

+ 20 

35. 

(+ 15) + ? = - 10 

45. 

(- 2 ) + ? = 

-20 


ADDITION AND SUBTRACTION OF POLYNOMIALS 51 


45. Rule for subtracting signed numbers. 


The rule for subtraction is taught 
by these four problems and their solu¬ 
tions in columns A and B 

(A) 

Sign op the 
Subtrahend 
as Given. 
Subtract 

(fi) 

Sign of the 
Subtrahend 
Changed. 

Add 

• 

n Example 1 . — Subtract + 2 



from -f 6 . 



This means (+ 2) + ? = + 6 . 

+ 6 

+ 6 

The result is + 4 (Column A). 

+ 2 

- 2 

This same result is obtained in Col- 

-f 4 

+ 4 

umn B by changing + 2 to — 2 and 



adding. 



Example 2. — Subtract — 2 



from + 6 . 



This means (— 2) + ? = + 6 . 

+ 6 

+ 6 

The result is + 8 (Column A). 

- 2 

+ 2 

This same result is obtained in Col¬ 

+ 8 

+ 8 

umn B by changing — 2 to +2 and 



adding. 



Example 3. — Subtract + 2 



from — 6 . 



This means (+ 2) + ? = — 6 . 

- 6 

- 6 

The result is — 8 (Column A). 

+ 2 

- 2 

The same result is obtained in Col¬ 

— 8 

- 8 

umn B by changing + 2 to — 2 and 



adding. 



Example 4. — Subtract — 2 



from — 6 . 



This means (— 2) + ? = — 6 . 

- 6 

- 6 

The result is — 4 (Column A). 

- 2 

+ 2 

The same result is obtained in Col¬ 

— 4 

-4 

umn B by changing — 2 to +2 and 



adding. 












52 


ALGEBRA 


Since the correct results are obtained in Column B, we have 
the following 

Rule. — To subtract one number from another, change the 
sign of the subtrahend and add the resulting number to the 
minuend. 

Example. — Subtract — 6 from — 13. 

Solution. — Mentally change — 6 to + 6; then add +6 — 13 

and — 13. The result is — 7. — 6 

— 7 

Check. The sum of — 6 and — 7 is — 13, as it should be. 

Historical Note. — The symbol -, like the symbol +, first ap¬ 
peared in print in a mathematical book by Widmann in 1489. The 
Italian and French mathematicians of the same period used the symbol 
m, derived from the first letter of the Latin word minus. 

EXERCISE 29 


Subtract the lower number from the upper number in each 
of the following examples : 


1 . 

2 . 

3. 

4. 

5. 

6 . 

7. 

8 . 

+ 16 

+ 20 

+ 12 

+ 9 

- 15 

- 13 

- 5 

- 8 

+ 7 

+ 13 

- 4 

- 5 

+ 8 

+ 4 

+ 7 

+ 10 

9. 

10 . 

u. 

12 . 

13. 

14. 

15. 

16. 

+ 8 a 

- 7 x 

+ 15 f 

- 13 1 

— 19 m 

+ 18 + 

-20x 2 

+ 14 

— 3 a 

+ 9 a: 

- Ty 2 

— 5 1 

— 6 m 

- 5 + 

+ 3 x 2 

- 8 


17. a. Was it possible to subtract 15 from 12 in arithmetic? 
b. What is the result in algebra when you subtract 15 from 12 ? 

18. a. Which is the subtrahend in a subtraction example? 
Which is the minuend? 

b. Is it necessary to write the subtrahend below the minuend ? 

19. a. Subtract — 5 from 13. 13 

b. Subtract 13 from — 5 without rewriting the numbers. — 5 
20-27. In Examples 9-16, subtract the upper numbers from 

the lower without rewriting the numbers. 
















ADDITION AND SUBTRACTION OF POLYNOMIALS 53 


23. From — 15 a 3 take — 23 a 3 . 

29. Subtract 18 abc from 3 abc. 

30. From —5 m 2 subtract — 8 m 2 . 

46. Subtraction of polynomials. 

Rule. — To subtract one polynomial from another: 

1. Rewrite the minuend, if necessary, in descending powers 
of one letter. 

2. Below it, write the subtrahend, so that like terms are in 
the same column. 

3. Imagine the signs of the terms of the subtrahend changed, 
and add the resulting terms to those of the minuend. 

Example. — Subtract 7 ab 2 — 9 a 2 b + 8 6 3 from — 2 a 2 b + 
4 ab 2 + 5 a 3 . 

Solution. — 1 . Arrange the terms in descending powers of a, with 
like terms in the same column. 

Minuend 5 a 3 — 2 a 2 6 + 4 ab 2 

Subtrahend — 9 a?b + 7 ab 2 + 8 b 3 

Change signs mentally, and add 5 a 3 + 7 a 2 b — 3 ab 2 —8 6 3 

Note. — In the first column, since nothing is added to 5 a 3 , the 
result is 5 a 3 . 

In the fourth column, + 8 b 3 becomes — 8 b 3 . Since nothing is 
added to it, the result is — 8b 3 . 

Check. — The sum of the remainder and subtrahend is the minuend, 
as it should be. 


1 . 

5a — 86 +11c 
2 a + 3 6 — 5c 

4. 

12 a 2 - 10 a - 6 
9 a 2 — 6 a + 5 


EXERCISE 30 
2 . 

11 x + lSy — 14 z 
4x + 9y — 16 z 

6 . 

4ic 2 + 13x — 10 
— 3 x 2 — x — 5 


3. 

2r + 7s - lOf 

2 r — 4s + 2 1 

6 . 

3 x 2 — 5 xy + y 2 
2 x 2 — 3 xy + 5 y 2 









54 


ALGEBRA 


7. i 

12 r 2 — 6 s 2 a 3 — 3 a 2 6 


8 . 


9. 



9 r 2 — 4 rg 4 2 s 2 a 3 4 3 a 2 6 4 3 ab 2 4 6 3 a; 4 — a: 2 ?/ 2 4 2/ 4 


10 . 11 . 

a? — 5 x 2 + 6 x y 3 +3?/—2 

2a: 2 -5 -y 2 + 4y - 1 


12 . 


a 3 - 3 a 2 +4 
4 a 2 — 2 a — 2 


Subtract: 

13. 5a — 6647c from 13 a —4648c 

14. 2 a: 2 — 3 a:?/ 4 4 ?/ 2 from 3 a: 2 — 5 xy — 6 y 2 

15. x 2 4 4 xy — 3 i/ 2 from 2 x 2 — b xy + y 2 

16. 4 a: 3 — 6 a: 2 4 8 x — 9 from 7 a? — 16 x 4 18 

17. ax — bx 4 cx from 2 ax — bx 4 cx 

18. — 3 z 4 11 x — 4 y from 15 a: — 5 y + 8 z 

19. 64-55430 from 84 425-70 

20. 2 a: 3 — 7 4 4 x — 5 x 2 from 6 a: 2 — 8 4 10 a: 3 — 3 x 

21. From l a? — 2 a + a 2 take 3 a — 2 4 a 2 — a 3 

22. Subtract 3 a: 3 — 544a: — 2 a: 2 from 7 a: 3 — 5 x 4 12 

- 5 a: 2 

23. Subtract 2 a: 3 — 3 x 2 44a: — 5 from 0 

24. Subtract - |n4Jp from fra — 4{p 

25. Subtract .5 x — .3 y 4 1-4 z from 2.8 x — .5 y 4 2.5 z 

26. From the sum of 3 a 2 — ab — 2 b 2 and 2 a 2 4 5 a6 — 3 6 2 
subtract a 2 — 3 ab — 4 6 2 

27. From 6 a: 2 - 7 a:?/ 4 8 ?/ 2 subtract the sum of a: 2 — 3 a:*/ 

— ?/ 2 and 2 a: 2 4 5 xy — 4 y 2 

28. From the sum of a 2 — 2 a6 4 6 2 and 2 a 2 4 2 a6 4 6 2 
take the sum of a 2 — b 2 and a 2 4 ab 4 3 6 2 

29. How much greater than ra 2 4 3 ran - n 2 is 4 ra 2 - 5 mn 

4 6 ?i 2 ? 








ADDITION AND SUBTRACTION OF POLYNOMIALS 55 


30. a. How much less than 3 a : 2 — 7 a: + 9 is 2 a : 2 + 4a: - 8 ? 

b. What is the value of the result when x = 2 ? 

Note. — Additional drill examples of the same kind are to be found 
in Exercise 218 on page 428, and some of a more difficult kind in Exer¬ 
cise 219 on page 429. 

47. Subtraction used in solving equations. 

After reading again the rules in § 11, § 12 , and §43, you 
easily understand the 

Rule. — The same number can be subtracted from both 
members of an equation without destroying the equality. 


Example 1. — Solve the equation 2 x + 17 =33. 

Solution. — 1. We can remove 17 from the left side of this equation 
by subtracting 17 from it. We must then subtract 17 from the other 
side also to keep the two sides equal. 

2x + 17-17 = 2z 

2. S 17 2 x = 16 33 - 17 = 16 

3. D2 x = 8 This can be done 

mentally. 

Check. — Substitute 8 for x in the original equation. 

Does 2 • 8 + 17 = 33? Does 16 + 17 = 33> 

Yes. Therefore 8 is the correct root of the equation. 

Note. — In Step 2, Si 7 means subtract 17 from both sides of the 
preceding equation. 


Example 2 . — Solve the equation 5 t + 6 = 2 t + 27. 

Solution. — 1. 5 t + 6 = 2 t + 27. 

2 . S 8 5 t = 2 t + 21 . 

3. S 2< 3* = 21 . 

4. D 3 t = 7. 

Check. — Substitute 7 for tin the original equation. 

Does 5 - 7+6 = 2 - 7+27? Does 35 + 6 = 14 + 27? 

Yes, since 35 + 6 = 41, and 14 + 27 = 41. 

Therefore 7 is the correct root of the equation. 

Note. — In Step 2 , 5 Z + 6 — 6 = 5 Z ; 2 t + 27 — 6 = 2 t + 21. 
In Step 3, 5 t - 2 t = 3 t ; 2 t + 21 - 2 t = 21. 

All this is to be done mentally. 



56 


ALGEBRA 


EXERCISE 31 

Solve and check the following equations: 


1. 

a: + 13 = 25 

8. 

10 x 4- 5 = 

= 13 + 6 * 

2. 

Sy + 11 = 32 

9. 

13 y + 3 = 

= 19 + 8 y 

3. 

5 t + 7 = 52 

10. 

7 s +8 = 

14+65 

4. 

4 k — 60 + k 

11. 

15 m + 7 

= 7 m + 55 

6. 

7 B = 28 + 3 B 

12. 

10 t + 9 = 

= 4 1 + 12 

6. 

9 r = 55 + 4 r 

13. 

20 w 4- 11 

= 15 w + 12 

7. 

12 z = 84 + 5 2 

14. 

18 x + 5 * 

= 15 a: + 7 

Solve the following equations, using the rules taught in § 43 

and § 

47: 




15. 

18y-7 = 3y+23 

23. 

15 6 + 4 = 

= 11-66 

16. 

3 5- 3= 2+ 2* 

24. 

32 IF - 6 

= 1 +4IF 

17. 

4y + 2 = Sy + 8 

25. 

10 x - 1 = 

= 6 a: + .6 

18. 

3 t + 9 = 37 — t 

26. 

8 y + 2 = 

3 y + 5.5 

19. 

5w-f-l=29— 2 w 

27. 

7r + 6 = 

8.2 - 4 r 

20. 

16 r - 13 = 4 r + 23 

28. 

4 c + 6 = 

12 - c 

21. 

12 £ + 5 = 12—2a: 

29. 

17 t - 4 = 

: 8 - 7 t 

22. 

15 6-4=3-66 

30. 

13 p + 5 = 

= - 2 p + 10 

48. 

Forming algebraic expressions 

for number relations. 


In arithmetic, you can actually find the sum of two numbers 
like 5 and 8. In algebra, you can only represent the sum of 5 
and x by writing 5 -f x. 

EXERCISE 32 

1. Represent 8 more than y ; 10 less than y. 

2. a. How much is 10 increased by 3 ? 

6. Represent 10 increased by n ; also x increased by 15. 

3. a. How much is 20 decreased by 9 ? 24 decreased by 18 ? 

b. Represent 15 decreased by x ; by m. 

c. Represent y decreased by 4; by 7. 


ADDITION AND SUBTRACTION OF POLYNOMIALS 57 


4. a. Represent m increased by 6. 

b. What is the value of the expression found in part a, when 
mis 4? 7? 10? 

6. A man spends y dollars out of his income of $200. 

a. How much has he left ? 

b. What is the value of the expression in part a, when y is 
25? 50? 100? 

6. a. How much does 12 exceed 9 ? 22 exceed 15 ? 

6. How much does 5 x exceed 3 a:? exceed 10 ? 

c. How much does y exceed 14 ? 20 exceed x ? 

7. The altitude of a certain rectangle is h inches. Its base 
is 4 in. longer. 

a. Represent its base. 

b. Represent its perimeter. 

8. One number is 4 times as large as a certain smaller num¬ 
ber. The smaller number is represented by s. 

a. Represent the larger number. 

b. Represent 15 more than the smaller number. 

c. Suppose the results of parts a and b are equal. Give the 
equation expressing this fact. 

9. One number is 6 times as large as a certain smaller num¬ 
ber. Let x represent the smaller number. 

a. Represent the larger number. 

b. Represent 20 more than the smaller number. 

c. Represent 5 less than the larger number. 

d. Suppose the results of parts b and c are equal. Give the 
equation expressing this fact. 

10. One number is 13 less than a certain larger number. Let 
g represent the larger number. 

a. Represent the smaller number. 

b. Represent t^ie sum of the two numbers. 

c. Suppose that the sum of the two numbers (represented in 
part b) is 88. Give the equation expressing this fact. 


58 


ALGEBRA 


Form equations for the following by direct translation, but 
do not solve the equations: 

Thus, A number decreased by 13 = 76 

Write n — 13 = 76 

11 . Three times a certain number equals 15 diminished by 
the number. 

12. Twice a certain number increased by 5 equals 18 dimin¬ 
ished by three times the number. 

13. If 9 be added to four times a certain number, the result 
equals 15 more than 3 times the number. 

14. Four fifths of a certain number increased by 16 equals 
the given number less 10. 

15. If 9 be added to 7 times a certain number, the result 
equals 16 diminished by the number. 

49. Solving problems by means of equations. 

A problem asks for values of one or more unknown numbers. 

Rule. — 1. Represent one unknown number by a letter. 

2. Represent the other unknown numbers of the problem by 
means of this same letter, using relations of these unknowns to 
the first unknown. 

3. Form an equation expressing an equality which is stated in 
the problem. 

a. Sometimes this equation expresses the fact that certain two re¬ 
sults are equal. 

b. Sometimes this equation expresses the fact that some expression 
containing the unknown number has a definite numerical relation to 
some known number. 

4. Solve the equation, thus getting the unknown number 
represented by the letter. Then find all the other unknown 
numbers required. 

Example. — One number is 9 times as large as another. If 
16 be subtracted from the larger, the result is 32 more than the 
smaller. What are the numbers ? 


ADDITION AND SUBTRACTION OF POLYNOMIALS 59 


Solution. — 1. Let $ = the smaller number. 

2. 9 s = the larger number. 

3. 9 s — 16 is 16 less than the larger number. 

4. s + 32 = 32 more’ than the smaller number. 

5. .*. 9 s — 16 = s + 32, because the problem states 

these two results are equal. 

6. An 9 s — s -f- 48. 

7. S 4 Ss = 48. 

8. D 8 s = 6, the smaller number. 

9. 9 s = 54, the larger number. 

Check. — If 16 be subtracted from 54, the result is 38. If 32 be 
added to 6, the result is 38. Since these two results are equal, and since 
54 is nine times 6, then 54 and 6 must be the correct numbers. 

Note. — Always check by going back to the statements made in the 
problem. 

EXERCISE 33 

1. What number increased by 45 equals 63 ? 

2. One number is 9 times as large as a certain smaller num¬ 
ber. If the smaller number is subtracted from the larger, the 
remainder is 32. Find the number. 

3. The result obtained by subtracting 24 from 7 times a 
certain number is 25. What is the number? 

4. If 12 be subtracted from 9 times a certain number, the 
result equals 7 times the number. What is the number ? 

6. There are two numbers of which the larger is 7 times the 
smaller. Also the larger is 48 more than the smaller. What are 
the numbers? 

6. One of two line segments is 4 times as long as the other. 
If 20 in. be added to the shorter, and 4 in. be taken from the 
longer, the two results will be equal. How long are the segments ? 

7. If 13 times a certain number be increased by 5, the result 
is 40 more than 8 times the number. What is the number ? 

8. If 3 be subtracted from 32 times a certain number, the 
result equals 15 more than 20 times the number. What is the 
number ? 




60 


ALGEBRA 


9. The length of certain lots in one city is 5 times their 
width. The perimeter of these lots is 300 ft. What are the 
length and the width of the lots ? 

10. The base of a certain rectangle is 3 times as long as its 
altitude. If the perimeter be increased by 10, the result is 50. 
What are the base and altitude of the rectangle ? 

11. The length of a certain parallelogram is 3 times its width. 
If the perimeter is increased by 6 ft., the result will be 110 ft. 
What are the length and width of the parallelogram ? 

12. 8 times a certain number exceeds (is more than) 3 times 
the number by 95. What is the number? 

13. 11 times a certain number exceeds 14 by 63. What is the 
number ? 

14. If 10 times a certain number be diminished by 15, the 
result is the same as if 6 times the number were increased by 21. 
What is the number ? 

15. Fifteen times a certain number diminished by 8 gives 
the same result as 6 times the number diminished by 5. What 
is the number? 

16. Eight times a certain number exceeds 37 by the same 
amount that 23 exceeds twice the number. What is the number ? 

17. The longest river in the world is the combined Mississippi- 
Missouri, and the next longest is the Nile. The length of the 
former is 700 mi. more than the length of the latter. Together, 
their length is 7700 mi. How long is each ? 

18. The area of the Caspian Sea, the largest salt-water lake, 
is 180,000 sq. mi. This area is 20,000 sq. mi. more than 5 times 
the area of Lake Superior, the largest fresh-water lake. What is 
the area of Lake Superior ? 

19. a . An article cost C dollars. Represent 15 % of this cost. 

b. Represent the cost plus 15 % of the cost. 

c. Suppose the result of part b is $2.99. Form the equation 
expressing this fact. 


ADDITION AND SUBTRACTION OF POLYNOMIALS 61 


d. Solve the equation formed in part c, and thus determine 
the cost of the article. 

20 . What is the cost of an article which sells for $7.44 if the 
gain is 24 % of the cost ? 

50. Most practical problems requiring an algebraic solution 
are solved by using a formula. 

The formula for the amount in simple interest problems is 
A = P + PRT because the amount equals the principal plus 
the interest, and the interest is PRT. In this formula, T is a 
number of years, and R is the rate of interest per year. 

Example. — At what rate of simple interest will $1500 amount 
to $2000 in 5 years ? 

Solution. — 1. The formula is A — P + PRT. 

A = 2000; P = 1500; R = ? T = 5. 

2. Substituting 2000 = 1500 + 1500 • R X 5. 

3. 2000 = 1500 + 7500 R. 

4. S 160 o 500 = 7500 R or 

7500 R = 500. 

5. D 7500 R = or 6 |-%. 

Check. — The interest on $1500 at 6$ % for 1 yr. = $100. 

/. the interest for 5 yr. = $500. 

.*. the amount = $1500 + $500, or $2000. 

.*• 6f% is the correct rate. 

Note. — Use the above form of solution in the following examples. 

EXERCISE 34 

1. If P represents the perimeter of a triangle whose sides 
have lengths a, b, and c, then 
! P = a b -f- c 
is the formula for the perimeter. 

a. Find P when a = 15, b = 40, c = 27. 

b. Find a when P = 67, b = 20, c — 30. 

c. Find b when P = 96, a = 22, c = 43. 

d. Find c when P = 72, a = 32.5, b = 16.8. 





62 


ALGEBRA 


2. The formula for the perimeter of a rectangle whose alti¬ 
tude is a and base is b is p_ G 

P = 2 a -f 2 b. 

a. Find P when a = 18 and b = 45. a 

b. Find a when P = 300 and b = 40. ^^ 

c. Find b when P = 275 and a = 46. 

3 . Z = a + (n — l)d is an important formula in mathe¬ 
matics. 

а. Find l when a = 5, n = 23, and d = 2. 

б. Find a when l = 68, n = 19, and <7=3., 
c. Find d when Z = 275, a = 25, and n = 26. 

4. Using the formula for the amount in simple interest 
problems: 

a. Find P when A = $3500, R — 5%, and T = 8 yr. 

b. Find R when A = $5000, P = $4000, and T = 2\ yr. 

c. Find T when A = $2600, P = $2000, and R = 6%. 


6 . 


By 


the formula S = 



a. Find S when n = 10, a = 13, and Z = 63. 

b. Find Z when S = 400, a = 5, and n = 40. 

c. Find a when 8 = 600, Z = 32, and n = 30. 


Note. — Additional problems of this sort are on page 430. Also 
attention is called to the tests on page 455. 




IV. PARENTHESES 


51. Parentheses are used to group together terms which are 
to be treated as a single number expression. 

Thus, 3 ct — (2 x y — z) 

means that 2 x + y — z is to be subtracted from 3 a. 

Brackets, [ ], braces, { J, and the vinculum, , are 

used in the same manner. For convenience, they are referred 
to collectively as parentheses, or as symbols of grouping. 

Historical Note. — Parentheses were introduced by Girard, a 
Dutch mathematician, about 1629. Previously, the brackets and the 
braces had been used by Vieta, about 1593, and by Bombelli, an Italian, 
about 1572. Descartes used the vinculum. 

52. Removing parentheses. 

Example 1. — Find the value of x 2 + (— x + 5). 

Solution. — x 2 + (- x + 5) means that - x + 5 is to be added 

X ‘ x 2 .'. x 2 + (— x + 5) 

- x + 5 = X 2 - x + 5. 

Adding, x 2 — x + 5 

Note. — The parentheses do not appear in the result; they have 
been removed. The sign of x in the result is —, just as it was inside 
the parentheses; also the sign of 5 is unchanged. 

Rule. — Parentheses preceded by a plus sign may be re¬ 
moved without changing the signs of the terms which are 
within the parentheses. 

Example 2. — Find the value of x 2 - (— x + 5). 

Solution .— x 2 - (— x + 5) means that - x + 5 is to be sub¬ 
tracted from x 2 . „ / • r\ 

x 2 .*. x 1 — (— X + 5) 

- X + 5 = X 2 + x - 5. 

x* + x - 5 


Subtracting, 


63 





64 


ALGEBRA 


Note. — The 'parentheses have been removed. The sign of x in the 
result is +, whereas it was — inside the parentheses; tRe sign of 5 in 
the result is —, whereas it was + inside the parentheses. 

Rule. — Parentheses preceded by a minus sign may be re¬ 
moved provided the signs of all terms which are within the 
parentheses are changed from -f to —, or from — to +. 

Example 3. — Remove the parentheses and collect like terms 
in 2 a — 6 b -f (4 a — b) — (5 a — 4 b). 

Solution. — 1. 4 a and — b, inclosed by parentheses preceded by a 
plus sign, do not have their signs changed when the parentheses are 
removed. 

5 a, which means + 5 a, and — 4 b, inclosed in parentheses preceded 
by a minus sign, must have their signs changed. 

2. 2 a - 6 b + (4 a - b) - (5 a - 4 b) 

= 2a — 6 6 + 4 a — b — 5 a + 4 h. 

3. Collecting like terms, the result is a — 3 b. 


EXERCISE 35 


Remove symbols of grouping and collect like terms: 


1. 3 m — (m -f 4) 

2. 4 x + (2 - 3 x) 

3. 6y - (5-4 y) 

4. 7t - (- St - 2) 

6. 8m+5-2m-5J 


6. 7 a + (— 2 a + 6) 

7. 6r - [- 5 +3r] 

8. 8c + \ — 3 c + 2 b] 

9. 9 k — [3 k - 1] 

10. 10 m - 5 - 3 n + 5m| 


Ex. No. 

Remove Parentheses 
FROM 

Find the Value of 
the Result 

11. 

3 to + (5 m — 2) 

when m = 2 

12. 

7 x - (4 x + 3) 

when x = 5 

13. 

5 1 - (6 - t) 

when t = 2 

14. 

2y + (-7+y) 

when y — 3 

15. 

32 — ( — 6 + z) 

when z = 4 









PARENTHESES 


65 


In the following ten examples, remove both sets of parentheses 
at the same time. Combine like terms. 


16. (4 c + 3 d) - ,(2 c - d) 

17. (x - 2 y) - (x + 3 y) 

18. (2 r + s) - (r - 4 s) 

19. [r + s] - [t + v] 

20. (x - y) - (2 y - x) 

53. Removing parentheses 
theses. 


21. \2a — 3b\ - 55a — 66j 

22. [— a -f- b] — [c — d\ 

23. (-o~6) + (-c-c0 

24. (a 2 — a) — (a 2 + a) 

26. (5 x — 8) — (3 x -f 4) 

which are within other paren- 


Example .— Simplify 6 x — (3 x -f \2 x — 4j — fa: — 5|). 
Solution. — The braces inclosing 2 x — 4 and x — 5 are removed 
first, giving attention only to the sign + before {2 x — 4} and the 
sign — before {x — 5}. The rest of the expression in merely rewritten. 

1. Qx - (3x -f {2x - 4} - {x - 5}) 

2. Removing {} = 6 x — (3 x 2 x — 4 — z + 5) 

3. C. T. inside () = 6 x — (4 » + 1) 

4. Removing (-) = 6 x — 4x — 1 

5. C. T. = 2x - 1. 


Note. — If x = 5, 2 x — 1 = 9. 
also has the value 9 when x = 5. 


Therefore the original expression 


Rule. — To simplify an expression having parentheses within 
parentheses: 

1. Remove the innermost parentheses first, according to the 
rules in § 52. 

2. Combine the terms within the resulting innermost paren¬ 
theses. Then remove these parentheses. 

3. Continue until all parentheses are removed and like terms 
are combined. 

EXERCISE 36 

1. 8 r + [5 r + (2 s — £)] 6. 6 r + [— 15 — (3 r — 9)] 

2. 3* + [7* •*-|2* + 1!] 6. 12 - }6a + (4 - 5*)} ' 

3. 9y + (2y-[8y-4]) \7.9f - {7 + (3«-2)L 

4. 11 2 + (- 4 z + [2 z - 5]) 8. 4 m - [3 m - (2 m + 1)] 



66 


ALGEBRA 


9. 6 w — f4 — (5 w — 8) j 

10. 5 A —[ — 2 A — (3 A — 1)] 

11. (7t - 5) - (3f - \2 t + 1J) 

12. (3 m — 4) — (—2 m -f- [5 m — 6]) 

13. (11 w - 5) - (- 4 - [6 w - 3]) 

14. 9r + 53 r — (6 r + 1) + (2 r - 8){ 

15. 7 z + (8 - [5 a; — 2] + [— 3 z + 4]) 

16. 6 6 - [4 6 - f2 + b\ - {3 b + lj] 

17. (4 t — 5) — (3 + \2 t - 7\ - J6 + 5 t\) 

18. 4x — |5z — [6z-b7a — 2]+5a{ 

19. 2 m - [(3 m - 4) - (5 m + 6)] 

20. 8 c - c - [2 c - 6] + [3 c - 7]| 

Note. — Supplementary examples are to be found on page 430. 

54. Placing terms inside parentheses. 

Development. — 1. What is the rule for removing paren¬ 
theses preceded by a plus sign ? 

2. What, then, should be done with the signs of terms which 
are placed within parentheses preceded by a plus sign ? 

Example 1. — Place the last two terms ofa + 6—c + din 
parentheses preceded by a plus sign. 

Solution. — a + 5 — c + d = a-\-b-\-{ — c + d). 

Check. — When the parentheses are removed 

« + & + (— c + d) becomes a + b — c + d. 

3. What is the rule for removing parentheses preceded by a 
minus sign? 

1. What, then, should be done with the signs of terms which 
are placed within parentheses preceded by a minus sign ? 

Example 2. — Inclose the last two terms of a' — b — c -f- d 
in parentheses preceded by a minus sign. 

Solution. — The signs of — c and + d must be changed. 

a-b-c + d = a- b-(+c-d). 







PARENTHESES 


67 


Check. — a — b — (c — d)=a — b — c + d when the parentheses 
are removed. 

Rule. — 1. If terms are placed within parentheses preceded 
by a plus sign, their signs must not be changed. 

2. If terms are placed within parentheses preceded by a 
minus sign, their signs must be changed. 

EXERCISE 37, 

Inclose the last three terms of the following expressions in 
parentheses preceded by a plus sign: 

1. x + y + 2 — w 6. a 2 — h 2 — 2 be — c 2 

2. r -f s — t + v 7. x 2 + y 2 — 2 yz + z 2 

3. m — n + p — g 8. 4m 2 — n 2 — Qn — 9 


9. x 2 — 4 ab — a 2 — 4 b 2 
10. m — nx — px — q 


4. a — h — c — d 
6. x 1 + x — 5 — y 


11 -20. Inclose the last three terms of the expressions in 
Examples 1-10 in parentheses preceded by a minus sign. 

21-30. Inclose the last two terms of the expressions in 
Examples 1-10 in parentheses preceded by a minus sign. 

65. Using parentheses in equations. 

Example. — Solve and check the equation 

(8 1 — 11) - (7 - 5 t) = 12 — (13 + 4 t). 

Solution. —1. (8 t - 11) — (7 — 5 t) = 12 - (13 + 4 t). 

2. Removing () 8«-ll-7 + 5* = 12-13-4*. 

3. C. T. 13 * - 18 = - 1 - 4 *. 

4 . A 18 13 t = 17 - 4 *. 

5. Au 17 * = 17. 

6 . D 17 t = 1. 

Check. — Substituting 1 for * in the original equation , 

does (8 - 11) - (7 - 5) = 12 - (13 + 4)? 

Does - 3 - 2 = 12 - 17? Does - 5 = - 5? Yes. 

1 is the correct root of the equation. 


68 


ALGEBRA 


EXERCISE 38 

Solve and check the following equations: 

1. 3 x — (x — 4) = 14 

2. (3 y-4) -(7+t/) =7 

3. 6 2 + (7 2 - 11) = 15 z - (4 z + 5) 

4. (4 ra - 13) - (5 + 2 ro) = (1 - 4 to) - (30 - 7 to) 

6. (10 1 + 3) - (8 1 - 7) = (15 - 6 f) - (14 1 - 6) 

6. (3 A — 4) — (2 A — 10) = (15 - 8 A) - (1 -5 A) 

7. (6 W - 5) — (9 W — 4) = (6 - 12 W) — (15 W - 1) 

8. (17 — 2 s) — (3 s — 15) + (4 s — 13) - 3 s - 21 

9. (5 r - 4) - (11 - 15 r) = (25 r - 3) - (10 r + 7) 

10. (4 c + 7) - (5 - 8 c) = 20 c - (12 c - 5) 

56. Using parentheses to represent number relations. 

Example. — Indicate and then simplify the excess of 7 x over 
the binomial (3 x — 4). 

Solution. — 1. “Excess” means “amount more than.” 

the excess of 12 over 9 is 12 — 9 or 3. 

2. Thus, the excess of 7 £ over (3 x — 4) = 7 x — (3 x — 4). 

3. Removing ( ) the excess = 7 x — 3x4-4, or4x + 4. 

EXERCISE 39 

Indicate and then simplify, if possible : 

1. The sum of x and (x + 7); of a: and (x — 5). 

2. (2 x + 5) decreased by (4 — x). 

3. (3 m — 8) decreased by (2 m + 5). 

4. The remainder when (2 £ — 3) is subtracted from 
(5 x + 6). 

5. The result when (3 + y ) is diminished by {y — 9). 

6. The amount by which (5 r — 4) exceeds (12 r + 3). 

7. The excess of 18 over x ; of 18 over (x — 5). 

8. The excess of (3 z + 4) over (2 z — 8). 




PARENTHESES 


69 


9. The smaller part of 18 if the larger part is l. 

10. The larger part of 20 if (x 3) is the smaller part. 

11. The number n is separated into two parts. If 10 is the 
larger part, what is the smaller ? 

12. A line segment of length 25 in. is separated into two 
parts, of which the shorter is s. Represent the longer part. 

13. a. 5, 6, and 7 are three consecutive integers. What are 
the two integers consecutive to them ? 

6. What must you do to an integer to get the consecutive 
integer ? 

c. If x is an integer, what is the first consecutive integer? 
The next? 

14. a. 2, 4, and 6 are consecutive even integers. What are 
the two even integers consecutive to them ? 

b. If x represents an even integer, what are the two consecu¬ 
tive even integers ? 

15. a. 7, 9, 11, 13 are four consecutive odd integers. What 
are the two odd integers consecutive to them? 

b. If x is an odd integer, what are the two consecutive odd 
integers ? 

c. What are these three integers if a? is 21 ? 

d. What are these three integers if £ is — 5 ? 

16. a. Let x = an integer. Then represent the two con¬ 
secutive integers. 

b. Now represent the sum of the three integers obtained in 
part a. 

17. Let a equal the altitude of a certain rectangle. 

a. If the base is 4 in. more than the altitude, represent the 
base. 

b. Represent the perimeter of this rectangle. 

c. How much is this perimeter when a = 5 ? 


70 


ALGEBRA 




18. The shortest side of a certain triangle is x in. long. 

a. Represent the second side if it is 8 in. longer than the first 
side. 

b. Represent the third side if it is 3 in. less than twice the first 
side. 

c. Represent the perimeter of this triangle. 


57. Using parentheses to solve problems. 

Example. — Separate 140 into three numbers such that the 
second is 15 less than the first and the third exceeds twice the 


first by 35. 

Solution. — 1. Three numbers are to be found. Their sum is 140. 
Each must be represented. 

2. Let / = the first number. 

3. .*. / — 15 = the second, since it is 15 less than the first. 

4. And 2/ + 35 = the third, since it exceeds twice the first by 35. 

5. •*./+(/— 15) + (2 / + 35) = 140, since the sum of the numbers 
is 140. 

6. Removing () / + / - 15 + 2/ + 35 = 140. 

7. C. T. 4/ + 20 = 140. 

8. S 20 4/= 120. 

9. D 4 / = 30, the first number. 

10. .'. / — 15 = 15, the second number. 

11. .\ 2/ -f 35 = 95, the third number. 
Check. — 1. The second is 15 less than the first, and the third exceeds 

twice the first by 35. 


2. Does 30 + 15 + 95 = 140? Yes. .*. 30, 15, and 95 are the re¬ 
quired numbers. 

Note 1 . — Use parentheses as in Step 5. 

Note 2. — Read Step 5 thus : 

/ plus the binomial / — 15 plus the binomial-2 / + 35 equals 140. 
Note 3. — Read again the suggestions given in § 49, p. 58. 


EXERCISE 40 

1. One number is 7 more than a smaller number. The sum 
of the numbers is 35. What are the numbers ? 

2. One number exceeds another number by 13. Their sum 
is 25. What are the numbers ? 




PARENTHESES 


71 


3. Separate 100 into two parts such that the larger is 14 
more than the smaller. What are the numbers ? 

4. Separate 124 into two parts such that the greater exceeds 
the smaller by 56. 

6. A boy has a board 36 in. long. He wants to cut it into 
two pieces such that one will be 8 in. longer than the other. 
How long will the pieces be ? 

6. Two angles, A and B, are complementary. (See Example 
4, p. 23.) Angle A is 21° smaller than angle B. How large is 
each? 

7. (Review Example 22, p. 25.) Let angles A, B, and C 
be the three angles of a triangle. If angle A is 50° less than 
angle B , and angle C is 5° more than angle B, how large is each ? 

8. The sum of the ages of three children of a certain family 
is 30 years. Charles’ age is 1 yr. more than twice John’s age; 
Mary’s age is 1 yr. less than 3 times John’s age. What are their 
ages? 

9. The base of a certain rectangle is 5 in. longer than the 
altitude. The perimeter of the rectangle is 66 in. Find the 
base and altitude. 

10. The perimeter of a certain triangle is 44 in. The second 
side is 1 in. less than twice the first side; the third side is 5 in. 
more than twice the first side. How long is each side ? 

11. The total height of the Library o f "Congress a t Washington 
is 64 ft. The height of its second story is 7 ft. less than twice 
the height of its first story; the height of its third story is 1 ft. 
more than twice the height of its first story. What is the height 
of each of its three stories ? 

12. There are two consecutive integers whose sum is 49. 
Wliat are they? (Review Examples 13 and 16, page 69.) 

13. There are three consecutive integers whose sum is 126. 
What are they ? 







72 


ALGEBRA 


14. There are two consecutive odd integers whose sum is 144. 
What are they? (Review Example 15, page 69.) 

15. (Review Example 15, p. 24.) Certain two angles are 
supplementary. One exceeds the other by 45°. How large is 
each? 

16. Separate 75.6 into two parts such that the larger is 49 
more than the smaller. 

17. Separate 55.5 into two parts such that the larger exceeds 
the smaller by 17.9. 

18. A board 14 ft. long is to be cut into two pieces such that 
one will be 1§ ft. longer than the other. How long will each 
piece be? 

19. Divide $175 between three children so that the first will 
receive $10 less than the second, and the third will receive $20 
more than the second. 

20. Charles’ age is 2 yr. more than twice John’s age. Henry’s 
age is 1 yr. more than three times John’s age. The sum of their 
ages is 33 years. How old is each? 

21. The base of a certain, rectangle is 4 in less than 5 times 
the altitude. The perimeter of the rectangle is 1 in. more than 
9 times the altitude. What are the base and altitude ? 

22. The sum of certain three consecutive integers is 204. 
What are they? 

23. If the first of certain three consecutive even integers is 
increased by the second, the sum is 20 more than the third 
integer. What are the integers ? 

24. Eight major planets, including the earth, revolve around 
the sun. The number of planets which are farther from the 
sun than the earth is one less than 3 times the number which 
are nearer to the sun than the earth. Find the number of 
planets nearer to, and the number farther from the sun than 
the earth. 




PARENTHESES 


73 


25. In a certain triangle, ABC, angle B is 20° more and 
angle C is 5° less than angle A. How large is each angle? 

26. Separate the total angular magnitude around a point 
into three angles such that the second is 45° more than the 
first, and the third is 45° more than the second. (See Example 
20, page 24.) 

27. Certain two angles are complementary. The larger ex¬ 
ceeds the smaller by 15. How large are the angles? 

28. The complement of a certain angle is 10° more than the 
angle itself. What is the angle ? 

29. The sum of the supplement and the complement of a 
certain angle is 150°. What is the angle ? 

30. There are three angles which make up the total angle 
around a point. The second exceeds 3 times the first by 45°; 
the third is 15° less than twice the first. How large is each ? 


V. MULTIPLICATION 


58. The law of signs for multiplication of positive and nega¬ 
tive numbers (§ 29, page 38) may be written: 

Rule. — 1. The product of two numbers having like signs is 
positive. 

2. The product of two numbers having unlike signs is negative. 

59. The law of exponents for multiplication. 

Development. — 1. Review the definitions of exponent , base , 

and power of a number, given in § 22, page 26. 

2. What does x 2 mean? a 3 ? r 7 ? 

3. Write in exponent form : 

(a) b • b • b • b (b) m •m • m • m • m (c) r • r • r • r • r • r 

(d) x • x • x ... x (if there are 11 factors) 

(e) y • y • y ... y (if there are 17 factors) 

4. The product of a 3 and a 4 is found as follows: 

a 3 = a • a • a.; a 4 = a • a • a • a. 
a 3 • a 4 = (a' a) X (a • a • a • a) = a • a • a • a • a • a • a = a 7 . 

5. Find, as in Step 4, the following products : 

(a) ic 2 • ic 4 = ? (6) r 3 • r 5 = ? (c) m 4 • m 5 = ? 

6. Observing the results in Step 5, get the following products 
mentally; check as in Step 4: 

(а) m 2 • m 3 = ? (c) z 2 • x 5 = ? 

(б) a 3 • a 5 = ? (d) 2 / 6 • 2 Z 3 = ? 

These examples teach the 

Rule. — To find the exponent of any number in a product, 
add the exponents of that number in the multiplicand and 
multiplier. 


74 


MULTIPLICATION 


75 


EXERCISE 41 

Using the above rule, give the following products: 

1. x 4 • z 3 4. t 4 • t 6 7. m 10 • m 3 10. 2 3 • 2 4 

2. t/ 6 • i/ 5 5. z 5 • z 6 8. s 9 • s 5 11. 3 2 • 3 3 

3. z 3 • z 7 6. fc 8 • k 4 9. w 8 • w 7 12. 10 3 • 10 5 

60. Multiplication of monomials. 

Development. — 1. How much is ( — 7) • (+2) ? ( — 5)*( — 4)? 

2. How does 9*5 compare with 5*9? Review §13, page 15. 

3. The product of — 5 x and — 6 x 2 y is obtained as follows : 

(- 5x)(- 6 x 2 y) = (- 5) • x • (- 6) • x 2 y 

— (— 5) • (— 6) • x • x 2 • y. Since the 
order of the factors may be 
changed. 

= + 30 x z y. 

4. Find, as in Step 3, the following products : 

(a) 3ran X2p; (6) (2a 3 6) X (-5a6); (c) (-7a) X (- 2ab). 

5. Give at sight, if possible, the following : 

(a) 2 rs • 5 t (c) 7 a • (— 3 a 2 ) 

(b) 3 xy • 2 z ((f) (-4m) *(-5 ran) 

Rule. — To find the product of two monomials : 

1. Find the product of the numerical coefficients, using the 
law of signs for multiplication. (§ 58.) 

2. Multiply this product by the literal factors, giving to each 
an exponent equal to its exponent in the multiplicand plus its 
exponent in the multiplier. 

Example. — Multiply - 5 x 3 yz 5 by + 6 xy 2 w. 

Solution. — (—5 3?yz h )(+ 6 xy*w) = — 30 xQ+Uy^+Vz^w 
= — 30 x^tfzPw. 


76 


ALGEBRA 


EXERCISE 42 


Find the following products : 


1 . 

a*- 

X 4 

16. 

(- 

9 m 2 n s ) (+ 12 mn 2 ) 

2 . 

(- 

2 x) • (+ 3 a:) 

17. 

(- 

12 xy 2 z) (+ 9 xy 2 ) 

3. 

(- 

4 y)(- by) 

18. 

(- 

6 a?b 4 ) (— 

a?b) 

4. 

(+*)(— 6 s 2 ) 

19. 

(+ 

8 ax 2 ) (— 

12 ax) 

5. 

(- 

w) (— 8 w 2 ) 

20 . 

(+ 

9 x 2 y 4 ) (— 

9 xy 2 ) 

6 . 

(- 

xy)(+ xz) 

21 . 

(- 

7 AB*)(+ 13 A 2 B) 

7. 

(- 

a 2 b)(— ab 2 c) 

22 . 

(- 

14 m 2 «)(- 

- n) 

8 . 

(+ 

mn 2 ) (— m 2 ) 

23. 

(- 

3 <*)(+# 

a 2 ) 

9. 

(- 

3 a6)(+ 2 a 2 ) 

24. 

(4n)(- f n 2 ) 

10 . 

(- 

±xy)\ - 6 x 2 y) 

25. 

(- 

12)(+f i 

r) 

11 . 

(- 

8 ab) (+ 3 a 2 ) 

26. 

(- 

15 xy){- 

fx 2 ) 

12 . 

(- 

3 r 2 s) (— 4 rs 2 ) 

27. 

(- 

40)(+|l 

') 

13. 

(+ 

5 xy 2 ) (—8 a: 2 ) 

28. 

(- 

iab) ■ (i 

a 2 ) 

14. 

<- 

11 rim?) (— 5 m 2 ) 

29. 

(- 

f »)(- 

f m 2 ) 

15. 

(- 

■ 6c*cP)(— 4 cd) 

30. 

(- 


;xy) 

31. 

a. 

What does the square of any number mean? 


b. Find : (a: 3 ) 2 ; (—5 a:) 2 ; (-f 2 xy) 2 ; ( — 3 a: 2 ) 2 . 

32. a. ^What is meant by the cube of a number? 

b. Find : (a: 2 ) 3 ; (-2 a:) 3 ; (- 3 a 2 ) 3 ; (- \ m)\ 


33. a. If the base of a certain rectangle is 5 a: in. and its 
altitude is 3 a: in., what are its area and perimeter? 

b. How much are the length, altitude, area, and perimeter 
of this rectangle when x is 2 ? 

34. The base of a certain triangle is 8 y in., and its altitude 
is 15 y in. What is its area ? 

35. a. The upper base of a certain 
trapezoid is 4 t in.; its lower base is 
1 It in.; its altitude is 6 t in. What is 
its area ? (Review § 23, page 28.) 

b. What is the value of this area when t is 10 ? 






MULTIPLICATION 


77 


36. The length of a certain rectangular parallelopiped is 
10 x in.; its width is 6 a; in.; its height is 4 z in. 

a. What is the area of its base ? 

b. What is the area of each of its side faces ? 

c. What is its volume ? 

37. The edge of a certain cube is 4 a; in. long. What is the 
volume of the cube ? 

38. What is the cost of m gallons of gasoline at p£ each ? 

39. What is the cost of 2 n tons of coal at c dollars each? 

40. The length of a certain rectangular parallelopiped is 5 x 
in.; its width is 3 y in.; its height is 4 z in. 

a . What is the area of its base' ? 

b. What is the area of one end ? 

c. What is the area of its front face ? 

Find: 

41. (.5 m) • ( — 3 m 2 ) 44. (— 6.8 t) • (+ .3 t 2 ) 

42. (- .8 x 2 ) • (+ .9 xy) 46. (f r 2 s) • (- is 2 ) 

43. (— 2.5 a 2 ) • (— 4 ab) 46. (— f- ran 3 ) • (+ m) 

MULTIPLICATION OF POLYNOMIALS BY MONOMIALS 
61. The product of two numbers may be rep- —| I | | | 
resented geometrically. 4 IIZII 

Example 1. —The product 5X4 may be repre- 11 

sented by the rectangle in Fig. 1. Notice that the 

. 0 _ ., Fig. 1 

area is 20 square units. 

Example 2. — The product 4(3 -f 7) may be represented by 

Fig. 2. - ■ ■ r . 


3 +■ 7 

Fig 



Check. — 


4(3 + 7) = 4 X 3 + 4 X 7. 
4 X 10 = 12 + 28. 















































78 


ALGEBRA 


Example 3. — The product a(b + c ) may be represented by 


Fig. 3 

a(b + c) = ab + ac. 

62. Multiplying polynomials by monomials. 

From the geometrical illustrations in § 61, it is clear that: 

a. 5(7 + 3) = (5 X 7) 4- (5 X 3) = 35 + 15 = 50 

b. 6(4 + 5) = (6 X 4) + (6 X 5) = 24 + 30 = 54 

c. a(b + c + d) = ab + ac + ad 

Rule. — To multiply a polynomial by a monomial, multiply 
each term of the polynomial by the monomial and combine the 
results. 

Example. — Multiply 3 a 2 — 2 ab + b 2 by — 3 ab. 

Solution. — 

(- 3 ab) X (3 a 2 — 2 ab + 6 2 ) = - 9 a 3 6 + 6 a 2 6 2 - 3 ab 3 . 
Check. — If a = 1 and 6 = 2, — 3 ab = —6; 

3 a 2 - 2 a6 + 6 2 = 3 - 4 + 4 = 3; and (- 6) X (3) = - 18. 
Also, the result, — 9 a 3 6 + 6 a 2 6 2 — 3 ab 3 , = — 18 + 24 — 24 = — 18. 

Note. — In this solution, (— 3 ab) (3 a 2 ) = — 9 a 3 6 ; (— 3 ab) ( — 2 a6) 

= + 6 a 2 6 2 ; and (— 3 ab) (6 2 ) = — 3 a6 3 . All this should be done 
mentally. 

EXERCISE 43 

Multiply: 

1. 5 a — 8 by 3 a 3» 7 xy{x —4 y) 

2. 3? - 4 x + 5 by 2 x 4. 6(-| x - |!/) 

See how rapidly you can obtain correct results for the follow¬ 
ing sixteen products. Write only the results. 

5. -4 z(3 x 2 + 2 x — 10) 7.-3 s(r 2 — 2 rs + s 2 ) 

6. 5 a 2 (2 a 3 - 5 a 2 + 7) 8.-3 z 2 (6z 4 - 7z 3 + 5z 2 ) 

9. - 2c 2 (- 2 c 2 + 3 cd + ld 2 ) 


ab 

ao 

> 



b + « 








MULTIPLICATION 


79 


10. — a?(x 5 — x* + 5 x 2 ) 12. x*y(a? -3 x*y + xy 2 ) 

11. — 6 ab( 10 a 3 — 5 6 2 ) 13. — rs(r 2 + rs -f s 2 ) 

14. — 5 ab(a 2 — 3 ab — 7 b 2 ) 

15. - 2 mn(4 m 2 — 3 mn -|- 2 n 2 ) 

16. 6 a 3 (3 a 2 - 4 a& + b 2 ) 

17. 5 m 2 -4mn + 3 n 2 19. 2 x 2 — 3 xy + 4 y 2 

2 to _ — 4 .r?/ _ 


18. 7 a 2 — 5 ab -j- 6 6 2 
— 3 ab 


20. z 3 — x 2 + 3 x — 4 
— 5 x 


21. Multiply § t + \ t — -J t by 60, collect terms, and find 

the value of the result when t = 2. 

✓o 1 o\ 12 o 30 i 20 n 

Solution. — 1. 60(-£ + =-t ~-t) = 00 • -1 + 00 • -t - 00 • - t 

\5 2 3 / 5 2 3 

2. = 36 < + 30 t - 40 t 

3. = 26 t. 

4. When t = 2, 26 t = 26 • 2, or 52. 

Multiply as in Example 21, and find the value of: 

22. 12^- — - + when x = 5. 

\2 3 4/ 

23. 20(p- - V -+ y -\ when y = 6. 

\2 5 4/ 

24. 15 (- — - 4* —^ when z = — 3. 

\5 3 15/ 

25. 30(^ t — T$t + it) when t = 10. 

26. 40(f x — ^ x — ^ x) when x = — 4. 

27. 36(-Jg w — w — J) when w = 8. 

28. 28(^ a + -§- b — \ c) when a = 2, 6 = 3, and c = 9. 

29. 24(§ x — £ y + i z) when x = 2, y = 1, and 2 = 5. 

30. a. The base of a certain rectangle is (a; -f- 8) in., and its 
altitude is 3 £ in. What is its area ? 

b. What is this area when x = 2 ? 






80 


ALGEBRA 


31. a. The altitude of a certain triangle is (y - 3) in., and its 
base is 12 y in. What is its area? 

b. What is this area when y = 5 ? 

32. a. There are n articles which cost ( n + 15) cents each. 
What is their total value ? 

b. What is this value when n = 10 ? 

33. a. A train travels t hours at the rate of (10 t -f 5) miles 
per hour. How far does it travel ? 

b. What is this distance when t = 4 ? 

34. Simplify 5 y — 3(x — 4). 

Solution. — 1. 3 times (x - 4) is to be subtracted from 5 y. 

2. But, by § 28, this is the same as adding — 3 (x - 4) to 5 y. 

3. .*. simply multiply {x — 4) by — 3 and write the results. Then 

5 y - S{x - 4) 

= 5 y - 3 x + 12. 


Simplify the following expressions, combining like terms. 


35. 5 y - 2(3 y - 4) 

36. 6 s + 8(4 s - 2) 

37. 7 t - 5(6 - St) 

38. - 8 w + 9(2 w - 4) 

39. - 13 x - 9(5 - 3 *) 


40. (3 2 — 7) — 6(4 z — 10) 

41. 9(11 -7 k) - 8(12 -5 k) 

42. 12(f m + 4) + 9(-J m — 5) 

43. 10(|-3 s) -8(4 -5 s) 

44. 16(fc + f) — 6(2 c — 3) 


63. Multiplying a polynomial by a polynomial. 

The product (x + y)(b + c) may be rep- & 
resented by the figure at the right, (x + y) 
is the base, and (b + c) is the altitude, c 
The areas of the four parts are marked. 

.*• (x + y)(b -f c) = bx + by + cx + cy 


bx 

by 

cx 

cy 


x y 


Rule. — To multiply one polynomial by another : 

1. Arrange multiplier and multiplicand in ascending or 
descending powers of the same letter. 

2. Multiply the multiplicand by each term of the multiplier. 

3. Add the partial products. 







MULTIPLICATION 


81 


Example. — Multiply 3a — 4 6 by 2a— 5 6, and check the 
solution by letting a = 1 and 6=2. 

Solution. — 3 a — 4 6 

2 a — 5 6 

6 a 2 — 8 ab This is 2 a X (3 a — 4 6) 

- 15 a6 -f 20 6 2 This is - 5 6 X (3 a - 4 6) 
Adding, 6 a 2 — 23 ab + 20 6 2 This is the product. 

Check. — a and 6 may be any numbers. 

When a = 1 and 6 = 2, 3 a - 4 6 = 3 - 8 = - 5.1 And (-5)(- 8) 
2a — 56 = 2 — 10= — 8. / =+40. 

The product, 6 a 2 - 23 a6 + 20 6 2 = 6 - 46 + 80; or 40. 

Since the two results are equal, the product is probably correct. 


EXERCISE 44 

Multiply: 

1. x + 9 by x + 7 

2. x + 11 by x + 13 

3. y — 7 by y — 5 

4. z — 8 by z — 10 
6. w + 4 by w — 12 

6. w — 16 by w + 3 

7. 2z + 5by3z + 2 

8. 6m— 3 by 4 m + 5 

9. 4 t — 5 by 6 t — 2 
10. 8 x — 3 by 5 x — 4 


11. 3 a + 2 6 by 2 a — 6 

12. 5m — 6/by3m+4£ 

13. 7w + 6pby8w— 4p 

14. 9 x 2 — y 2 by 2 z 2 + 3 y 2 
16. 3 w 2 — 7 m by 4 w 2 + 5 m 

16. § a + 6 by 2 a + 6 

17. ' | y by \ x + £ ?/ 

18. %x — 3zby-f:r— 3z 

19. fm-f*byfm-f* 

20. iz + iyby^z-^i/ 


21. (z 2 + 2 x — 3)(z — 1) 

22. (x 2 + xy + 2/ 2 ) (a - y ) 

23. (z 2 - zy + 2 / 2 )(x + y) 

24. (m 2 — 4 m + 3)(m — 2) 

25. (a 2 — 2 a — 4)(2 a + 3) 

26. (ic 2 - 3 zy + 2 y 2 ){x —2 y) 

27. (3 r 2 + 6 r -8)(4r+ 3) 




82 


ALGEBRA 


28. (4 m 2 — 2 mn + n 2 )(m — 3 n) 

29. (4 s 2 + r 2 — 4 rs)(r — 2 s ) 

30. (3 c 2 + d 2 — 2 cd)(2 c -f d) 

31. (4 n 2 - 7 + 5 n)(n 2 - 2 n + 1) 

32. (5* + 2* 2 - 6) (3 t 2 - 5* + 4) 

33. (a 2 - 3 - 2 a) (a 2 + 3 - 2 a) 

34. (5 x — 7 + 4 x 2 ) (3 +2 x 2 - 3 x) 

35. (9 r 2 + 4 s 2 -f 6 rs)(9 r 2 + 4 s 2 — 6 rs) 

36. ( x 2 - 2 xy + ?/ 2 )(x 2 + 2 xy + y 2 ) 

37. (x 3 - x 2 i/ + xy 2 - i/ 3 )(z + 2/) 1 

38. (a 3 + 2 a 2 b + 4 afr 2 + 8 6 3 )(a —2 b) 

39. (r + s - t)(r - s + t) 

40. (2 a — 3 b + c) (2 a — 3 b — c) 

41. (3x - 2 y) 2 43. (4 r - 5 s) 2 45. (x -f i/) 4 

42. (x — 3 yf 44. (2 m - 3 n) 3 46. (x - y ) 4 

47. a. What is the area of the rectangle whose base is (5 x + y) 
in., and whose altitude is (2 x — y) in. ? 

b. How much is this area when x = 3 and y = 4 ? 

48. a. What is the volume of the cube whose edge is (2 a — b) 
in. ? 

6. What is this volume when a = 4 and 6=3? 

49. a. What is the cost of (2 n + m) articles at (n — m)ft 
each? 

6. How much is this result when n = 8 and m = 3 ? 

60. a. A man traveled (n + 8) hr. at the rate of (m + 2) mi. 
per hour. How far did he go ? 

b. What were his rate, time, and distance when m = 30 and 
n — 3? 

Note. — Supplementary examples in multiplication are on page 431. 


MULTIPLICATION 


83 


64. Multiplication used in solving equations and problems. 

Example. — Seven times the complement of a certain angle 
exceeds twice its supplement by 20°. Find the angle. (Review 
Examples 4 and 15, page 23.) 

Solution. — 1. Let a = the number of degrees in the angle. 

2. Then 90 — a = the number of degrees in its complement, 

3. and 180 — a = the number of degrees in its supplement. 

4. 7(90 - a) - 2(180 - a) = 20. 

5. Multiplying, 630 — 7a — 360 + 2 a = 20 . 

6 . C. T. - 5 a + 270 = 20. 

7. S 270 - 5 a = - 250. 

8 . M_! 5 a = 250. 

9. D 5 a = 50, the number of degrees 

in the angle. 

Check. — The complement of the angle is 40°, and the supplement 
is 130°. Does 7 X 40 - 2 X 130 = 20? Does 280 - 260 = 20? 
Yes. 

Note. — In Step 8 , both members are multiplied by — 1 to change 

— 5 a into +5 a. 

In Step 5, 180 multiplied by — 2 gives — 360; — a multiplied by 

— 2 gives +2 a. (Review Example 34, page 80.) 

You will need to take similar steps in the following examples. 


EXERCISE 45 


Solve and check the following equations : 


1. 3(s - 4) = 33 

2. 2(3 y - 8) = 20 

3. 5(2 x - 1) = 35 

4. 6(8 x - 1) = 4x + 16 
6. 7(1 - 9 0 = 12 t - 18 
6. 3(1 —12 m) = 4 m — 7 

13. 3(2 x — 4) — 2(x — 5) = 2(x + 6) 

14. 6(2 - 3 y) + 30 = 3(4 y - 6) 

15. 14 - 5(3 a - 2) = 3(o - 10) 


7. 4(y - 3) - 5 = 3 

8. 6(z + 2) - 40 = 50 

9. 8(3 t - i) + 9 = 17 

10. 2(3 r - 11) = 5(11 - r) 

11. 4(16 - m) = 3(m - 9) 

12. 2(7 x + 1) = 3(7* - 4) 


84 


ALGEBRA 


16. 10 - 5(r - 4) = 6(¥ - r) 

17 . 2(9 - 3 v) + 3(4 - v) = 21 

18. - 8) - 6(2 g - 15) = 4(6 - g) 

19. 3(5 - n) - 4(2 n - 7) + 3(7 + n) = 0 

20. 4(3 m - 5) + 3(6 m - 1) = 3(9 m - 7) - 1 

21. The greater of certain two numbers is 7 more than the 
smaller. Twice the greater number is 6 more than three times 
the smaller. Find the numbers. 

22. One number exceeds a certain other number by 5. Six 
times the smaller number, diminished by four times the larger 
number, is 6. What are the numbers ? 

23 . Separate the number 35 into two parts such that twice 
the larger, diminished by three times the smaller, is 10. 

24 . The sum of certain two numbers is 45. If four times the 
smaller be decreased by twice the larger, the result is 12. What 
are the numbers ? 

25 . One number exceeds another number by 3. Six times 
the larger, diminished by seven times the smaller, is 7. What 
are the numbers ? 

26 . There are two consecutive integers such that the sum of 
twice the larger and four times the smaller is 110. What are 
these integers? 

27 . There are certain two consecutive integers such that the 
square of the larger minus the square of the smaller gives 25. 
Find these integers. 

28 . There are three consecutive integers such that three times 
the first, plus two times the second, plus the third, gives 100. 
Find these integers. 

29 . There are three consecutive even integers such that the 
first, plus twice the second, plus three times the third, gives 100. 
What are they ? 


MULTIPLICATION 


85 


30. The base of a certain rectangle is 8 in. more than the side 
of a certain square ; its altitude is 3 in. less than the side of the 
square. The perimeter of the rectangle is 74 in. What is the 
length of a side of the square, and what are the base and altitude 
of the rectangle ? 

31. If 5 times the complement of a certain angle be subtracted 
from 2 times its supplement, the result is 15°. What is the 
angle ? 

32. If the supplement of a certain angle be subtracted from 
four times its complement, the remainder is 5° more than four 
times the angle itself. What is the angle ? 

EXERCISE 46 

1. A is now 18 years old. Express his age : 

a. 6 years ago; b. n years ago; c. x years from now. 

2. B is x years old. Express his age : 

a. 4 years hence (from now); b. 6 years ago; c. y years ago. 

3. A is now x years old. B is 8 years older. Express : 

a. B’s present age; b. the sum of their ages; 

c. the age of each 10 years ago; 

d. the age of each 5 years hence. 

4. A is now n years old. B is four times as old. Express : 

а. B’s present age; b. the age of each 4 years hence; 

c. the fact that B’s age as found in part 6/is 3 times A’s age. 

d. Solve the equation formed in part c. Obtain the age of 
each. 

б. B is three times as old as A is now. Make a table and 
supply the answers to the questions asked below. 

a. Let a = the number of years in A’s age now. 



Now 

10 Years Ago 

A’s age 

? 

? 

B’s age 

? 

? 







86 


ALGEBRA 


6. Write the equation which expresses the fact that B’s age 
ten years ago was 5 times A’s age then. Solve it, and obtain 
the ages of A and B. 

c. Check the solution by finding their ages ten years ago, and 
comparing them with the statement in the problem. 

Solve the following problems : 

6. A is now four times as old as B. In six years he will be 
twice as old as B is then. How old is each now ? 

7. C is now five times as old as D. In fifteen years he will 
be twice as old as D is then. How old is each? 

8. A is now 12 years older than B. Four years from now, 
A will be twice as old as B is then. What are their present ages ? 

9. The sum of the ages of a father and son now is 30 years. 
In 9 years the father’s age will be only 3 times the son’s age 
then. How old is each now ? 

10. A’s age is now 38 years and B’s is 5 years. 

a. How old will each of them be in x years ? 

b. If A will be 4 times as old as B at that time, form the 
equation from the ages obtained in part a, and solve it for x. 

11. A’s age is now 33 years, and B’s is 3 years. In how many 
years will A’s age be just 6 times B’s age at the same time ? 

12. A’s age now is 65 years, and B’s is 20 years. How many 
years ago was A 6 times as old as B was ? 

Equations for problems may often be written at sight. 

Thus, for Example 6 we can write at once: 

(46 + 6) = 2(6 + 6). 

Write at sight the equations for the following problems, and 
solve them as rapidly as you can. Time yourself. 

13. A is four times as old as B now. Five years ago, he was 
7 times as old as B was then. Find their ages. 

14. A is 5 times as old as B. Five years ago, he was 4 times 
as old as B will be one year hence. Find their ages. 


MULTIPLICATION 


87 


16. A is now 6 years older than B. Three years ago, A was 
twice as old as B was then. Find their ages. 

16. A is now 10 years older than B. Five years from now 
he will be 5 times as old as B was five years ago. Find their 
ages. 

17. A is now 7 times as old as B. A’s age in four years will 
be 5 times what B’s age will be in two years. Find their ages. 

EXERCISE 47 

Mixture and Coin Problems 

1. A grocer has some tea and some 40^ tea. How many 
pounds of each must he take to form a mixture of 100 pounds, 
which he can sell for 60^ per pound ? 

Solution . — 1. Let n = the number of pounds he must take, 

of the 90^ tea. 

2. .'. 90 = the value of these n pounds. 

3. Then 100 — n = the number of pounds he must take, 

of the 40^ tea. 

4. .*. 40(100 — n)£ = the value of these (100 — n) pounds. 

5. .*. 90 n + 40(100 — n) = the total value of the tea. 

6. Also 100 X 60^ = the total value of the tea. 

7. .*. 90 n + 40(100 - n) = 100 X 60. 

Complete the solution, obtaining n. Answer the questions asked in 
the problem, and check your solution. 

Note. — The essential statements in Steps 2-6 can be arranged thus : 


Kind of Tea 

No. of Pounds 

Value in Cents 

90^ tea 

40^ tea 

60^ tea 

n 

100 - n 

100 

90 ni 

40(100 - n)i 

100 X 60 


Many teachers advise pupils to substitute this table for Steps 2-6 of 
the solution above. 

2. A grocer wishes to make a mixture of 60 pounds of coffee 
to sell at 40^ per pound. He has some 55^ coffee and some 30^ 
coffee. How many pounds of each should he take ? 









88 


ALGEBRA 


3. A seedsman wishes to make a mixture consisting of clover 
seed and timothy seed which he can sell at 11^ per pound. His 
clover seed is worth 23 4 per pound and his timothy 9^ per pound. 
How many pounds of each should he take for a batch of 200 
pounds ? 

4. A grocer has 20 pounds of 60^ coffee. How many pounds 
of 30^ coffee should he mix with it so that the mixture can be 
sold at 35^ per pound ? 

6. A dealer has walnuts valued at 43^ per pound and almonds 
valued at 35^ per pound. How many pounds of each should he 
take to make a mixture of 150 pounds which he can sell at 40^ 
per pound ? 

6. A sum of money amounting to $2.80 consists of pennies, 
nickels, and quarters. The number of nickels is 3 times the 
number of pennies, and the number of quarters is 3 more than 
the number of pennies. How many of each coin are there ? 

Solution. — 1. Let p = the number of pennies, 

2. then pfc = their value in cents. 

3. .*. 3 p = the number of nickels, 

4. and 15 pi = their value in cents. 

5. .*. p + 3 = the number of quarters, 

6. and 25 (p + 3)i = their value in cents. 

7. p + 15 p + 25 (p + 3) = the total value of the coins in cents. 

8. Also 280 = the total value of the coins in cents. 

9. .'. p + 15 p + 25 (p + 3) = 280. 

Complete the solution, obtaining p, and check your solution. 

Note. — The essential statements in Steps 2-8 can be arranged thus : 


Kind of Coin 

Number of Them 

Value of Them 
in Cents 

Pennies 

V 

V 

Nickels 

3 P 

15 p 

Quarters 

P +3 

25 (p + 3) 

All together 


280 


Many teachers advise pupils to substitute this table for Steps 2-8 of 
the solution above. 









MULTIPLICATION 


89 


7. A sum of money amounting to $2.55 consists of nickels 
and dimes. The number of dimes is 6 more than the number 
of nickels. How many of each kind of coin are there ? 

8. A sum of money amounting to $5.35 consists of nickels, 
dimes, and quarters. There are 5 more quarters than there 
are nickels; the number of dimes is 1 less than three times the 
number of nickels. How many of each kind of coin are there ? 

9. A sum of money amounting to $38.75 consists of quarters, 
half-dollars, $1.00 bills, and $2.00 bills. The number of half- 
dollars is 2 more than the number of quarters ' the number of 
$1.00 bills is 3 less than twice the number of quarters ; the num¬ 
ber of $2.00 bills is 1 less than the number of quarters. How 
many of each kind of coin are there ? 

10. A girl’s savings bank had in it pennies, nickels, dimes, 
and quarters. Their total value was $10.16. The number of 
dimes was twice the number of quarters. There were 15 more 
nickels than there were dimes. The number of pennies was 3 
less than four times the number of quarters. How many of 
each kind of coin were there ? 

65. Equations having fractional coefficients. 

Example. — If the sum of twice a certain number and one 
half of the number be diminished by two thirds of the number, 
the result is 27§. What is the number ? 

Solution. — 1. Let n = the number. 

2. Then 2n + ^n-|n = 27£. 

3. To obtain an equation which does not have in it any fractions, 
multiply bv the smallest number which contains all the denominators as 
divisors. This is called the least common multiple of the denominators. 

M 6 6(2 n + hn -■£ n) = 6 X ^ 

4. .'. 12 n + 3‘■n -* 4 n — 165. 

5. C. T. * 11 n = 165. 

6. Dn n — the number. 

Check. — Twice the number = 30; one half the number =7^; 

two thirds the number = 10. Does 30 + 7^ — 10 = 27^? Yes. 


^X—= 165 
- 2 " 


X—ft = 4 n 



90 


ALGEBRA 


6 . 


EXERCISE 48 

Solve the following equations : 

1 x x — 9 
2 5 ” 10 

2. y+ y -=™ 

3 3 

3 2a: . x - 9 
5 4 10 

A ht \ 2t 

4 - y~3 = T 


a: _ _9_ , x 
4 _ 10 ' 5 


9. 


3 s 
8 

4 w 

T" 

5 m 

IT 


s_ = 2 
3 3 

2_w 
3 

26 _ 2 m 
5 ~ 5 


= —+ 2 


6- ^ = - + 5 

3 4 3 

10. 

X _ X 1 

5 4 2 

11. !l + *r_2l-7 

2 4 3 

16. 

. c _ 62 c 

7 7 3 

12. t — — = 10 — — 

4 6 

17. 

a = 12 - - - - 

3 6 

13 c_13_c_3c 
’3 6 2 5 

18. 

3s 7 _ 4s 

2 3 5 

14. 3d-— =-+ ^ 

6 2 9 

19. 

2 x — — — 1 2 3 4 ^ . 
3 2 

j j 7 m _ m _ 3 3m 
'8 5"i ¥ 

20. 

3 w 4 w 4 

~5~ ~S ~ 9 “ 


J7 

16 


EXERCISE 49 

1. Two thirds of a certain number is 20 more than one 
fourth of the number. What is the number ? 

2. What number is such that eight fifths of it, decreased by 
50, gives a remainder of fourteen fifteenths of it? 

3. What number increased by one eighth of itself equals the 
sum of 30 and three sixteenths of itself ? 

4. One half a certain number equals the excess of five 
sevenths of the number over 6. What is the number ? 


MULTIPLICATION 


91 


6. What number exceeds by 13 the sum of one third itself, 
one fourth itself, and one fifth itself ? 

6. There are three consecutive integers such that the sum 
of the second and third exceeds one fourth of the first by 17. 
What are these integers ? 

7. There are three consecutive integers such that the sum 
of the second and third is 21 more than one fifth of the first. 
What are these integers ? 

8. The perimeter of a certain triangle is 41 in. The second 
side is nine sevenths of the first side; the third side is one half 
the second side. How long is each side ? 

9. The perimeter of a certain quadrilateral is 47 in. One 
side is one fourth of the longest side; a second side is two fifths 
of the longest; the third is seven tenths of the longest. How 
long is each side ? 

10. There are four angles whose sum is the total angle around 
a point. (See Example 20, page 24.) The second angle is five 
sevenths of the first; the third is one half of the first; and the 
fourth is one half of the second. How large is each angle ? 

11 . Certain two angles are complementary. (See Example 
4, page 23.) The smaller is two sevenths of the larger. How 
large are these angles ? 

12. Certain two angles are supplementary. (See Example 
15, page 24.) The smaller is seven elevenths of the larger. 
How large is each of these angles ? 

13. In certain triangles having three unequal angles, the 
smallest angle is one sixth of the largest, and the remaining 
angle is one third of the largest. How large is each of the 
angles ? (See Example 22, page 25.) 

14. In other triangles, the smallest angle is three tenths of 
the largest, and the remaining angle is one half of the largest. 
How large is each of the angles ? 


92 


ALGEBRA 


16. The perimeter of the Treasury at Washington, D. C., is 
1400 ft. Its width is five ninths of its length. How long and 
how wide is it ? 

16. A’s age is now two thirds B’s age. A’s age 4 years from 
now will be five sixths of B’s present age. How old is each now ? 

17. A’s present age is one third of B’s. A’s age 10 years 
from now will be what B’s age was 10 years ago. How old is 
each now? 

18. A man has $4.50 in nickels, dimes, and quarters. He 
has three fourths as many dimes as he has nickels, and five 
eighths as many quarters as he has nickels. How many coins 
of each kind has he ? 

19. The total value of a sum of money consisting of nickels, 
quarters, and half-dollars is $16.35. The number of half- 
dollars is two thirds of the number of nickels. The number 
of quarters is four thirds of the number of half-dollars. How 
many of each coin has he ? 

20. A grocer has a mixture of two kinds of coffee, having a 
total value of $49.50. One kind was worth 30j£ per pound, 
and the other, 45^ per pound. The number of pounds of the 
latter used in the mixture was one fourth of the number of 
pounds of the former. How many pounds of each did he use ? 


VI. DIVISION 


66. Division is the process of finding one of two numbers 
when their product and the other number are given. 

To divide 15 by 3 means to find the number by which 3 must be 
multiplied to give the product 15. 

The dividend is the product of the numbers; it is the number 
divided. 

• The divisor is the other given number; it is the number by 
which the dividend is divided. 

The quotient is the required number. 

67. It is clear that a -f- a = 1; for a X 1 — a. 

68. Division by zero is impossible. 

Thus, if we try to find the quotient of 6 -f- 0, and let q equal the 
quotient, we should have the relation 

6 = 0 • g. 

But 0 • q = 0 and not 6, so there cannot be any number to use as q. 
Hence, there is no number to represent the quotient of 6 -f- 0. 

69. Division is indicated by writing a fraction whose numera¬ 
tor is the dividend and whose denominator is the divisor. 

Thus, the quotient of 15 -5- 5 is written 

The quotient of 7 abc -s- 3 xy is written ? — • 

o xy 

Note. — The line — was used to indicate division long before the 
symbol -5-. 

70. Division of a product by a number. 

Example. — Divide 6 X 8 by 2. 

Solution. — 

1. If 6 is divided by 2, = 1^2 = 24. 


93 



94 


ALGEBRA 


2. If 8 is divided by 2, = 24. 

2 2 

3. If both 6 and 8 are divided by 2, 

3 4 

6X8_0XB_ 1O 
2-2- 12 - 

4. Since we know that ^ X = 48 4- 2 or 24, it is clear that the re- 

2 

suits obtained in Steps 1 and 2 are correct, but that the result in Step 3 
is incorrect. Hence, 

Rule. — To divide the product of two or more numbers by a 
number, divide any one of the factors by the number, but divide 
only one of them by it. 


EXERCISE 50 

Find each of the following indicated divisions in two ways : 

1 9 X 12 18 X 24 28 X 56 

3 '6 *7 

71. The law of signs for division. 

Since (+2) X (+ 3) = +6, then (+6) 4- (+2) = +3. 

Since (—2) X (+ 3) = — 6, then (—6) -4 (— 2) = +3. 

Since (+ 2) X (— 3) = — 6, then (— 6) 4- (+ 2) = — 3. 

Since (—2) X ( — 3) = +6, then (+ 6) 4 ( — 2) = — 3. 

If the signs of the dividend, the divisor, and the quotient in 
each of the previous statements are examined, the following 
rules become clear: 

Rule. — 1. The quotient of two numbers having like signs is 
positive. 

2. The quotient of two numbers having unlike signs is nega¬ 
tive. 











DIVISION 


95 


EXERCISE 51 


Find: 

1. (+ 18) * (+ 3) 

2. (+ 12) -s- (+ 2) 

3. (+ 15) 4- (+ 5) 

4 . (- 30) 4- (- 6) 

5. (-45) -s- (-9) 

6. (-50) *f- (-5) 

7. (+ 18) 4- (-2) 

8. (- 24) * (+ 3) 

9. (-36) 4- (-6) 

10. (+42) - (- 7) 

11. (- 100) 4- (- 20) 

12. (+72) 4- (- 12) 

13. (— 81) -s- (+ 3) 

14. (- 90) 4- (- 10) 

15. (-84) 4- (+7) 

16. (+ 60) 4- (- 10) 

17. (+ 90) 4- ( - 30) 

18. (+ 6) 4- (— 12) 

19. (- 10) 4- (-20) 

20. (- 5) 4- (+ 4) 


21. 

(- 

15) - 

-(- 

•15) 

22. 

<+ 

32) - 

-(- 

■ 32) 

23. 

(- 

18) - 

-(- 

■1) 

24. 

(+ 

17) - 

-(- 

-1) 

25. 

(- 

108) 

-( 

-3) 

26. 

(+ 

120) 

-( 

-8) 

27. 

(- 

33) - 

-(- 

- ID 

28. 

(- 

64) - 

(+ 16) 

29. 

(+ 

80) - 

-(- 

-5) 

30. 

(- 

39) - 

-(- 

- 13) 

31. 

(+ 

16) - 

-(- 

- 16) 

32. 

(- 

91) - 

-(- 

-7) 

33. 

(- 

200) 

4- (+ 5) 

34. 

(+ 

6) 4- 

(- 

18) 

35. 

(- 

7) - 

(- 

35) 

36. 

(- 

4) -H 

(+ 

6) 

37. 

(- 

5) 4- 

(- 

8) 

38. 

(- 

9.6) 

+ (■ 

-3) 

39. 

(+ 

5;8) 

- ( 

-2) 

40. 

(- 

6.4) 

-( 

-2) 


72. The law of exponents for division. 

Development. — 1. Review the definitions of exponent, base, 
and power of a number in § 22. 

2. Divide o 5 by a 3 . 

Ill 

c , .. a® <? ■ n -a ■ u ■ a _ , „ _ . 

Solution. -, =-= 1 ■ a ■ a — a‘. 

a 6 o ■ a - a 

111 

Therefore a 8 4* a 3 = a 2 . Check. — a 3 • a 2 — a 8 . 

Each a in the denominator is divided into one of the a’s in the numera¬ 
tor. The quotient in each case is 1, since a 4- a = 1. 









96 


ALGEBRA 


3. Find as in Step 2 the following quotients : . 

a. y 6 -f- y* = ? b. m 6 -5- m 4 = ? c. f 10 -s- f 8 = ? 

4. Examine carefully the exponents in the dividend, the 
divisor, and the quotient. In the following problems, try to 
give the results immediately without going through the solu¬ 
tion as in Step 2. Test by multiplication. 

а. p s -r- p 6 = ? c. b 7 -7- 6 4 = ? 

б. fl «vfl 5 =? d. c 8 -j- c 4 = ? 

5. Divide a 4 6 6 by a 3 6 2 . 

Ill 11 

SoJuft'on. — a 4 6 6 + aW = ^ = ^ ' a -ft #- b- b • b ■ b = a!) , 

a 3 5* ^ ^ ^ M 

11111 

Rule — The exponent of any number in the quotient is equal 
to its exponent in the dividend minus its exponent in the divisor. 

Historical Note. — This rule was known to Stifel. 

Notice that b 2 -s- b 2 = b 2 ~ 2 = 6° by the law of exponents. 

Since b 2 4- b 2 must equal 1, we agree that b° = 1. 

The zero power of any number is 1. 

Thus: a? = 1; 5° = 1; c° = 1. 

73. Division of monomials by monomials. 

Example 1. — Divide 54 a 7 6 2 c 3 by —9 a 4 6 2 c 2 . 

Solution. — ■ 54 = t- 6 a (7—4)6(2-2) c (3-2) 

- 9 aW 

= — 6 a 3 6°c, or — 6 a 3 c. 

Check. — (— 6 a 3 c) • (— 9 <i 4 b 2 c 2 ) = + 54 a 7 b 2 c?. 

Rule. — To divide a monomial by a monomial: 

1. Make the quotient positive, if the dividend and divisor 
have like signs ; make it negative, if they have unlike signs. 

2. Find the quotient of the absolute values of the numerical 
coefficients. 

3. Multiply the quotient of Step 2 by the product of the literal 
factors, giving each its exponent in the dividend minus its ex¬ 
ponent in the divisor. 




DIVISION 


97 


4. Omit any literal factor which has the same exponent m the 
dividend and divisor. 

Example 2. — Divide — 33 a 6 bx 2 y 4 by + 3 a b x 2 y. 

Solution. — (—33 a*bx 2 y A ) (+ 3 a b x 2 y) = — 11 aby 3 . 

Check. — These solutions may be checked by substitution, for they 
must be correct for all values of the literal numbers (except 0 some¬ 
times). A better way is to use the rule that the divisor times the 
quotient equals the dividend. 

Here, does (+ 3 a b x 2 y) X (— 11 aby 3 ) = — 33 a?bx 2 y A ? Yes. 

EXERCISE 52 

Divide: 


1. y 6 by y 4 

2. m 7 by m 2 

3. r 8 by r 5 

4. t 9 by f 4 

6. a 3 6 2 by ab 

6. rV by r^s 

7. x 6 y 3 by x 2 ?/ 3 

8 . m 4 n 6 by m 4 n 

9. x b y 3 by x 4 y 3 

10. m 3 n 2 by mn 2 

11. 12 a 3 by 3 a 2 

12. 18 x 4 by 6 x 2 

13. 20 rV by 5 r 2 s 

14. 28 c 4 ^ by 7 c 2 ^ 

15. — 14 mn by — 7 n 

16. + 18 a 2 6 by — 3 ab 

17. — 12 xy 3 by — 6 xy 2 

18. — 24 iPt by + 8 r 2 

19. — 34 m 2 n 3 by + 17 mn 3 

20. + 40 x b y 6 by — 8 x 3 !/ 2 


21. 

— 32 a?b 4 c by 

— 4 a6 4 c 

22. 

35 x 5 i/V by — 

7 x 3 ?/ 2 

23. 

— 44 a b xy z by 

— 4 a 4 y 2 

24. 

75 x 1 V* 8 by - 

15 x?/ 3 

25. 

42 aW by - 

14 ah 2 <? 

26. 

— 64 x 8 y 6 by - 

- 8 x 7 y 6 

27. 

— 42 ab s c b by 

- 7 6 2 c 4 

28. 

- 60 xVz 6 by 

- 12 x 2 y*z 4 

29. 

— 60 a 4 6 5 c 3 by 

- 20 a 4 6 3 

30. 

+ 84 c 5 d 6 by - 

- 12 c 4 d 

31. 

- 51 a6 3 c 2 by 

— 3 ac 2 

32. 

+ 48 r 5 s 6 by - 

-16 r 4 s 6 

33. 

5 ab by 10 a 


34. 

4 mn 2 by 8 m 


35. 

— 3 r by — 15 

I r 

36. 

-f 9 cd by — 12 c 

37. 



38. 

(—3 «o + (! 

c) 

39. 

3 5 4- 3 4 


40. 

2 6 4- 2 3 



98 


ALGEBRA 


74. Division of polynomials by monomials. 


Development. — 1. Since 2 X 9 = 18, then ^ 8 - = 9. 

2. Since 2{x + 3) = 2 x -f- 6, then " x ^ = x + 3. 

3. Since 3(a — 5) = 3 a — 15, what does —-— equal? 

O 

4. What does each of the following equal? Test the result 

by multiplying the divisor and quotient. 

2x 2 , 6 r + 4 6m — 9 

a. - - —; b. - ! —; c. ---- 

2 2 3 

5. Since a(b + c) = ab -f- ac, 

Then - + - " 0 = b + c. 

a 

Rule. — To divide a polynomial by a monomial: 

1. Divide each term of the dividend by the divisor. 

2. Unite the results with their signs. 

Example 1. — Divide 12 .r 3 — 6 x 2 + 3 x by — 3 x. 

Solution.- 12.x 3 — 6a^4-3x _ _ 4x * + 2x - 1. 

~ 3 X 

Check. — (— 4 z 2 + 2 x — 1) • (— 3 x) = 12 x 3 — 6 x 2 + 3 x. 

Example 2. — Divide — 9 a 3 + 3 a 2 — 12 a 4 by — 3 a 2 . 
Solution. — (- 12 a 4 - 9 a 3 + 3 a 2 ) (- 3 a 2 ) = 4 a 2 + 3 a - 1. 

Check. — Multiply the quotient by the divisor; the result should 
equal the dividend. 


Divide: 

1. 5 r — 10 s by 5 

2. 12 p — 8 q by 4 

3. 15 x 2 — 18 y 2 by 3 

4. 9 x* — 7 x by x 


EXERCISE 53 


6. 18 m 3 — 12 m by 3 m 

6. 6 x 4 - 3 z 3 by - 3 x 

7. — 20 r 3 + 14 i^s by — 2 r 2 

8. 12 m 2 n — 18 mn 2 by — 6 mn 


9. — 33 a 2 b + 44 ab 2 by — 11 ab 
10 . 28 c 3 d - 21 cd 3 by - 7 cd 









DIVISION 


99 


11. - 72 r 6 - 80 r 4 + 64 r 2 - 32 r by - 8 r 

12. 15 z 3 - 18 x 2 y + 21 xy 2 by - 3 z 

13. - 48 z 10 + 96 z 8 - 72 z 6 by - 12 z 5 

14. 12 y 5 — 8 y 4 + 16 y 3 by — 4 y 2 

16. 60 ra 9 — 45 ra 6 + 30 m 3 by — 15 m 3 

16. 33 a 3 6 3 c 3 — 44 a 2 6V + 55 abc by — 11 abc 

17. 25 ra 7 n 6 — 30 ra 3 n 4 + 40 ra 2 n 3 by — 5 ran 2 

18. 15 c 4 d 2 — 20 c 3 d 3 — 25 c 2 d 4 by — 5 c 2 d 2 

19. 6 a 2 6 2 c 3 — 9 a 3 6 3 c 4 + 12 a 4 6c 2 by — 3 a 2 bi? 

20. — 49 x 3 y 4 z 2 + 56 x 2 yh A by — 7 x 2 y 2 z 2 


21. 

a 5 b — a z b z + ab b 

24. 

— 2 ra 2 + 4 ran 

— 6 n 2 

ab 

- 2 


22. 

x* - 12 x 3 + 6 x 2 

26. 

16 x 3 y — 12 x 2 y 2 

+ 8z?/ 3 

z 2 

-4 y 


23. 

— ra 2 + ra — 1 

26. 

— 5 ra 3 + 10 ra 4 

_ 9 

— 15 ra 5 


- 1 

75. Division of polynomials by polynomials is like long 
division in arithmetic. 


Example 1. — Divide 864 by 24. 

36 

Solution. — 1. 86 -s- 24 = 3 + . 24 864 

2. 24 X 3 = 72; subtract . _7?_ 

3. 14 -5- 2 = 6+ 144 

4. 24 X 6 = 144; subtract . 144 

Example 2. — Divide 10 z 3 — 21 z 2 — 11 z + 12 
by 2 z 2 — 3 z — 4. 

Solution . — 5 x — 3 _ 

1. 10 z 3 ^ 2 z 2 = 5 x. 2 z 2 - 3 x - 4 

2. (2 z 2 - 3 x — 4) X (5 x) ; subtract . 

3. (-6Z 2 ) -i- (2 x 2 ) = -3. 

4. (2 z 2 - 3 x - 4) X (- 3) ; subtract . 


10 z 3 — 21 z 2 — 11 * + 12 
10 z 3 - 15 z 2 - 20 z 


- 6 z 2 + 9 z + 12 

- 6 z 2 + 9z + 12 















100 


ALGEBRA 


The following explanation of the process may be given. 

We are to find an expression which, when multiplied by the divisor, 
2 x 2 — 3 x — 4, will produce the dividend. 

The term containing the highest power of x in the product is the 
product of the terms containing the highest powers of x in the multi¬ 
plicand and multiplier. 

Therefore, 10 x 3 is the product of 2 x 2 and the term containing the 
highest power of x in the quotient. Dividing 10 x 3 by 2 x 2 gives 5 x, 
which is the term containing the highest power of x in the quotient. 

When the dividend is formed, the divisor is multiplied by each term 
of the quotient, and the results are added. Now reversing the process, 
multiply the divisor by 5 x and subtract the result, 10 x 3 — 15 x 2 — 20 x, 
from the dividend. 

The remainder, — 6 x 2 + 9 x + 12, must be the product of the 
divisor and the rest of the quotient. Consider it a new dividend. 

Its term containing the highest power of x, — 6 x 2 , is the product of 
2 x 2 and the term containing the next lower power of x in the quotient. 
Dividing — 6 x 2 by 2 x 2 gives the next term, — 3. Multiply the divisor 
by — 3 and subtract the result from the previous remainder. There is 
now no remainder. 

The quotient is therefore 5 x — 3. 

Rule. — To divide a polynomial by a polynomial: 

1. Arrange the dividend and the divisor in either ascending 
or descending powers of some common letter. 

2. Divide the first term of the dividend by the first term of 
the divisor, and write the result as the first term of the quotient. 

3. Multiply the whole divisor by the first term of the quotient; 
write the product under the dividend and subtract it from the 
dividend. 

4. Consider the remainder a new dividend, and repeat Steps 
1, 2, and 3. 

Note 1. — The terms of the quotient are placed above the terms of 
the dividend from which they are obtained. 

Note 2. — Like terms are arranged carefully in a vertical column. 

Note 3. — Spaces should be left for any powers of the common 
letter which are not present in the dividend. 

Note 4. — As in arithmetic, there may be a final remainder. 


DIVISION 


101 


Example 3. — Divide x 2 y 2 + x 4 - y 4 by - xy + y 2 + x 2 . 


Solution. — 


x 2 -xy + y 2 


x 2 + xy + y 2 _ 

x 4 + x 2 y 2 - y 4 

x 4 — x?y + x 2 y 2 _ 

+ X?y - y 4 

+ x?y — x 2 y 2 + xy 3 _ 

x 2 y 2 — xy 3 — y 4 

x 2 y 2 - xy 3 + y 4 
- 2 y 4 


Note. — The quotient is x 2 + xy + y 2 ; the remainder is —2 y 4 . 
As in arithmetic, the complete quotient may be written: 

Complete quotient: x 2 + xy + y 2 + ■ ■■ ~ 2 - 

x 2 — xy + y 2 

Check. — Let x — 1, and y — 1 . Then, dividend = 1 , and divisor 

- 1 . 

Quotient = 1 + 1 + 1 + -— ^ = 3 + =3 — 2 = 1 . 

Since 1X1 = 1, the quotient is correct. 

Another check would be to multiply the divisor by the quotient and 
add the remainder; the sum should equal the dividend. 


EXERCISE 54 


Divide: 

1 . x 2 -f 5 x + 6 by x + 3 

2 . x 2 + 8 x + 12 by * + 2 
3- y 2 + 8 y + 15 by y + 5 

4. m 2 + 13 m + 12 by m + 1 

5. A 2 + 10 A + 24 by A + 6 

6. r 2 - 18 r + 32 by r - 2 

7. s 2 - 12 s + 35 by 5 — 7 


8 . t 2 + 72 - 17 t by t - 9 

9. c 2 + 72 — 18 c by c 6 

10 . cP + 13 d + 36 by d + 4 

11 . z 2 + 2 x - 8 by x - 2 

12 . t 2 + t - 30 by t + 6 

13. a 2 — 6 a — 27 by a — 9 

14. m 2 + to — 20 by m + 5 


15. n 4 — 5 n 2 — 24 by n 2 + 3 

16. x 2 — 16 ax + 55 a 2 by x —5a 

17. a 2 — 3 ah — 108 h 2 by a + 9 h 

18. x 2 — 2 xz — 48 z 2 by x — 82 

19. x 4 + 6 x 2 y — 27 y 2 by x 2 —3 y 









102 


ALGEBRA 


20 . 

21 . 

22 . 

23. 

24. 

25. 

26. 

27. 

28. 

29. 

30. 

31. 

32. 

33. 

34. 

35. 

36. 

37. y 3 

38. y 3 

39. x 4 

40. a 4 
46. 

46. 

47. 


x 2 y 2 — 18 xy + 45 by xy — 3 
15 z 2 — 14 x — 16 by 3 x + 2 
6 a 2 -f 25 — 25 a by 2 a — 5 
24 a 2 — 38 a + 15 by 6 a — 5 
30 x 2 + 7 — 47 x by 6 x — 1 
28 x 2 — 15 y 2 + 23 xy by 4 x + 5 y 
20 m 2 + 41 mn + 20 n 2 by 5 m + 4 n 
a^ — 5 x 2 — 17 x -f* 66 by x — 6 
6 a 3 — 27 a — a 2 + 20 by 3 a — 5 

4 a? 3 — 5 xy 2 —|— 4 i/ 3 — 8 a^y by 2 a? — y 

12 a 3 + 17 a 2 6 — 25 6 3 — 20 a& 2 by 3 a + 5 b 
x 4 — x*y — 2 x 2 y 2 — 3 xy 3 — y 4 by z 2 + xy + y 2 

5 n 3 + 8 ?i 2 — 23 n - 1 by 5 n 2 — 7 n — 2 

15 — 30 ic — 8 — 19 a: 2 by - 5 x + 3 x 2 - 4 

2 a 4 + 3 a 2 - 8 a 3 + 7 + 10 a by 2 a 2 - 7 - 4 a 
4 m 3 - 3 m - 15 m 2 + 4 by m 2 - 3 m - 3 
2 a^ — 9 a; 2 y + 17 xy 2 — 12 y 3 by 2 a; — 3 y 
+ 27 by y + 3 41. 8 a 3 — 1 by 2 a — 1 

— 27 by y + 3 42. a; 4 — 16 y 4 by a; — 2 y 

— y 4 by £ — y 43. 27 m 3 + 64 n 3 by 3 m -f 4 n 

-f 6 4 by a + b 44. a 5 + b b by a + b 

16 n 4 - 81 by 8 n 3 + 12 n 2 + 18 n + 27 

6z 4 — 7a^ — 28x 2 + 8a: — 21by3.T 2 — 5z — 7 
x 4 + x 2 y 2 + y 4 by x 2 + xy + y 2 


Historical Note. — Stifel (1486-1567) seems to have been one of 
the first to divide a polynomial by a polynomial. Sir Isaac Newton 
(1642-1727), in a book published in 1707, pointed out the advantage 
of arranging the dividend and divisor according to ascending or descend¬ 
ing powers of the same letter. 

Note. — Supplementary examples appear on page 431. 


VII. SIMPLE EQUATIONS 

76. An identity is an equation which is satisfied by all values 
of the literal numbers except possibly zero. 

It is a declarative sentence. It is a statement of actual 
equality under all conditions. 

Thus, 3 x(a — b) = 3 ax — 3 bx is an identity. 

If a = 3, b = 1, and x = 2, 3 x{a — b) = 3 • 2(3 — 1) = 6 • 2 or 12. 

Also, 3 ax - 3 bx = 3 • 3 • 2 — 3 • 1 • 2 = 18 - 6 or 12. 

Try any other values of a, b, and x. You will find that they also 
satisfy this equation. 

Similarly, (3 a + 2 b) + (a — b) is identically equal to 4 a — b 
15 x 2 — 3 x 2 is identically equal to 12 x 2 

7 x{2 y — z) is identically equal to 14 xy — 7 xz 

30 y 3 -T- 3 y is identically equal to 10 y 2 

That is, the result obtained from adding , subtracting, multiplying, or 
dividing two expressions is identically equal to the indicated combination 
of the numbers. It is for this reason that you can check solutions of 
such problems by substitution of any numbers for the letters. 

77. A conditional equation is an equation involving one or 
more literal numbers, which is not satisfied by all values of the 
literal numbers. 

Thus: a. x + 2 = 5 is not satisfied by any value of x except x = 3. 

b. x 2 — 5 x — — 6 is satisfied by x = 2 and by x = 3, but by no 

other values of x. 

A conditional equation is like an interrogative sentence; it 
implies a question. 

Thus, 3 x — 5 = 4, asks “for what value of x is 3 x — 5 = 4?” 

The answer is, “x must be 3,” for 3X3 — 5 = 9 — 5 = 4. 

The word “ equation ” usually refers to a conditional equa¬ 
tion. 


103 



104 


ALGEBRA 


78. If an equation has only one unknown number, if the 
unknown does not appear in the denominator of any fraction, 
and if the unknown appears only with the exponent 1, then the 
equation is called an equation of the first degree, or a simple 
equation. 

Thus, 3x - 5 = 4 is a simple equation. 

Historical Note. — The idea of the degree of an equation was 
introduced by Descartes. 

Properties of Equations 

79. You have learned four rules to be used in solving simple 
equations. If you do not recall these rules, review §§ 11, 12, 
43, and 47. 

There are three mechanical methods of solution which can 

he used, although use of them is unnecessary. 

80. Transposition. 

Example. — Solve the equation 10 £ —5 = 3 £ + 30. 

Solution. — 1. 10 x — 5 = 3 x + 30. 

2. Ag 10 a; = 3 z + 30 + 5. 

3. S 33 10 z — 3 x = 30 -f 5. 

4. C. T. 7 x = 3£ 

5. D 7 x = 5. 

By adding 5 to both members of Step 1 , — 5 is made to disappear 
from the left member and + 5 to appear in the right member in Step 2 . 
It is as if — 5, with its sign changed, were taken from the left side and 
placed on the right side. 

In Step 3, subtracting 3 x from both members of Step 2 , causes 3 x 
to disappear from the right side of Step 2 , and — 3 x to appear in the 
left side in Step 3. Again, it is as if 3 x, with its sign changed, were 
moved from the right side to the left side. 

These are two examples of transposition (“place across”). 

Rule. — A term may be transposed from one member of an 
equation to the other, provided its sign is changed. 

Historical Note. — Our word algebra, curiously, is associated with 
this process, transposition. About the first quarter of the ninth cen- 


SIMPLE EQUATIONS 


105 


tury an Arabian mathematician, Mohammed ben Musa, wrote an 
algebra, for the title of which he used Ilm al-jabr wa'l muqabalah. Al- 
jabr meant the process of transposing terms. This title was used in 
various forms in Europe until about the fifteenth century, when the 
last part was dropped and algebra came into use. 

The Greeks had no special name for their algebra. The Hindu 
writers called it reckoning with unknowns. 

81. Canceling terms in an equation. 

Example. — Solve the equation x + a = b + a. 

Solution. — 1. x + a = b + a. 

2. S a x — b. 

Notice that the term + a, which appeared in both members of the 
given equation, has disappeared from both members in Step 2. 

The result in Step 2 is the same as if + a were simply dropped from 
both members. It was not simply dropped; it was subtracted from 
both members ., However, 

Rule. — If the same term, preceded by the same sign, occurs 
in both members of an equation, it may be canceled (dropped) 
from both members. 

Example. — Solve the equation 

5 x 9 x — 8+3 = 9 # + 18 — 8 + 2 x. 

Solution. — 1. 5x+0*-8+3 = M + 18-8+2*. 

2. Since + 9 * appears in both members, it can be canceled; also 
— 8 can be canceled. 

.*. 5 * + 3 = 18 + 2 x. 

3. Transposing 

+ 2 * and + 3, 5 x — 2 x = 18 — 3. 

4. C. T. 3x= 15. 

5. D 3 x — 5. 

Check. — Substituting 5 for * in the original equation , 
does 25 + 45 - 8 + 3 = 45 + 18 - 8 + 10? 

Does 73 - 8 = 73 - 8? Yes. 

Note. — First, cancel any terms which appear in both members; 
second, transpose terms containing unknowns to the left side, and 
terms containing knowns to the right side. 



106 


ALGEBRA 


EXERCISE 55 

Solve the following equations : 

1. 4m-2=m + 7 4. 5?/—3 = 11+2?/ 

2. 8 f + 5 = 3 t + 35 5. 7 a - 11 = 15 - 6 z 

3. 7 x — 10 = 50 — 5 x 6. 15 r — 3=4 — 6 f 

7. 13 w — 92+8 = 4 + 5 w + 8 

8. 12fif-5fif + l= 8fir + 3- 5^ 

9. 11a — 14 + a = 9a + 4 + a 

10. 20 s - 19 s + 3 = - 19 s + 13 - 5 $ 

11. 4z — 7 + 5z — 9 = 61+5# — 3z 

12. 32 A + 16 + 14 A + 1 - 9 A = 7 A + 16 + 14 A + 13 

13. 17m-fl6ra—5ra—4 = 5m+8—3ra + 2 + 16m 

14. 141 — 13 + 11* - 17 -2< =8* - 13 — 16* + 25 + 11 1 

15. 8 & + 6.5 - 48 - 73 k = 48 + 5 k + 6.5 - 17 k - 73 k 

82. Changing signs in an equation. Often we obtain an 
equation like 

_ x + 8 = + 3. 

We prefer that the sign of x be plus. To make it so, multiply 
both members by — 1. Then, 

x — 8 = — 3. 

The sign of each term now is exactly opposite to its sign in 
the original equation. The result of multiplying by — 1 is the 
same as if the signs were simply changed. 

Rule. — The signs of all terms of an equation may be changed, 
without destroying the equality. 

83. Negative roots of equations. 

Example. — Solve the equation 

7 — 5x — 9z = 15— 9z — 3x. 


SIMPLE EQUATIONS 


107 


Solution. — 1. 7 — 5 x — 9 x = 15 — 9 x — 3 x. 

2. Canceling — 9 x, 7 — 5x=15 — 3x. 

3. Transposing + 7 

and — 3 x, 3 x — 5 x = 15 — 7. 

4. C. T. - 2 x = 8. 

5. D _2 x = - 4. 

Check. — Substitute — 4 for x in the first equation. 

Does 7 - 5(— 4) - 9(- 4) = 15 - 9(- 4) - 3(- 4)? 

Does 7 + 20 + 36 = 15 + 38 + 12? 

Does 63 = 63? Yes. 

Note. — In Step 5, (- 2 x) + (- 2) = x; (+ 8) + (- 2) * - 4. 
Or, you may multiply first by — 1 in Step 4, and then divide by + 2. 


EXERCISE 56 

Solve the following equations either by the rules taught in 
§ 11, 12, 43, and 47, or by the ones taught in §§ 80, 81, and 82. 


1. 2 — 3 x = x — 8 

2. 4m — 5 = 6m + 1 

3. 13 — 6x = 3 — 11 x 

4. 7 t + 10 = 19 t - 2 
6. 12 — 5n = n + 36 


6. 11 + 7 v = 3p - 13 

7. 9 r — 5 = 4r + 5 

8. 23 — 7 2 = - 15 2 + 15 

9. 16 + 3 w = — 2 w + 6 

10. 19 - 16 y = 24 y - 1 


11. 3(m + 1) - 9 = 5(2 m + 7) - 6 (Review § 64, page 83.) 

12. 8 t - 5(4 t + 2) = 3(5 - 3 t) - 1 

13. 8 - 3 x(x - 4) = 2 - x(3 x - 15) 

14. 6(3 - 2 c) - 5 c = 5 c + 7(4 - 3 c) 

15. 3(3 x + 7) — 4(6 x 3) = 10 6(3 x 4) 

16. 10 r - (4 r + 3) = 3(8 r - 2) + 4(1 - 4 r) 

17. m(m — 7) — 20 = 2 m(m — 11) — m(m - 13) 

18. 6(4 x -3) = 5 - 10(3 - x) 

19. 4(3 r - 7) + 1 = 5(6 r - 1) - 2(5 - 9 r) 

20. 19 - 6(4 c + 1) = 5(8 c + 1) - 12(4 c - 1) 

84. Solution of problems. You have been taught how to 
form the equations for all the problems of the following list, 




108 


ALGEBRA 


excepting one kind. Example 1 is one of this new kind. In 
it, you must express the relation between a dividend, a divisor, 
the quotient, and the remainder. 

If 19 is divided by 7, the quotient is 2 and the remainder is 5. 

Then 19 = 2 • 7 + 5. 

That is, dividend = quotient X divisor + remainder. 


EXERCISE 57 

1. a. Write the equation expressing the relation between 
dividend, divisor, quotient, and remainder when 46 is divided 
by 8. 

b. Select a dividend and a divisor, and write the equation 
connecting your dividend, divisor, and the quotient and re¬ 
mainder you obtain. 

2. If (37 — x ) is divided by x, the quotient is 3 and the re¬ 
mainder is 5. Form the equation expressing this relation, and 
solve it for x. 

3. If n is divided by 65 — n, the quotient is 3 and the re¬ 
mainder is 5. Find n. 

4. Separate 54 into two parts such that the larger part 
divided by the smaller part gives 3 as quotient and 10 as re¬ 
mainder. 

Hint. — What is the dividend in this example? The divisor? ~ 

5. Separate 125 into two parts such that the larger divided 
by the smaller gives 3 as quotient and 21 as remainder. 

6. If 13 be added to a certain number, and the sum be 
multiplied by 2, the product is 60. What is the number ? 

7. Separate 28 into two parts such that 5 times the smaller 
exceeds 4 times the larger by 5. 

8. Separate 43 into two parts such that 2 times the larger, 
diminished by 4 times the smaller, equals 7 less than the larger. 


SIMPLE EQUATIONS 


109 


9. There are three consecutive integers such that 4 times 
the first, diminished by the second, equals 12 more than 2 times 
the third. What are they? 

10. What are the three sides of the triangle whose perimeter 
is 121 in., if the second side is twice the first, and the third side 
is 19 in. shorter than the second ? 

11. The Lincoln Memorial at Washington stands on a plat¬ 
form whose perimeter is 676 ft. The length of the platform 
exceeds its width by 70 ft. What are the dimensions of the 
platform ? 

12. The foundation of the Lincoln Memorial goes down to 
the rock below the level of the Potomac. From the rock below 
grade to the top of the memorial is 172 ft. If the height above 
ground be divided by the depth below, the quotient is 2 and 
the remainder is 22. What is the depth below grade, and what 
is the height above grade ? 

13. The age of A is now 4 times the age of B. Four years 
from now A’s age will be- one year more than 3 times B’s age 
then. What are their present ages ? 

14. John is 4 years older than Charles. Eight years ago, 
twice Charles’ age exceeded John’s age then by 1 year. How 
old is each now ? 

15. The treasurer of a club, after collecting dues, had $17.00 
in nickels, dimes, quarters, and half-dollars. The number of 
dimes was 3 times the number of nickels ; the number of quarters 
exceeded the number of dimes by 2 ; the number of half-dollars 
was 3 less than 5 times the number of nickels. How many of 
each coin did he have ? 

16. The total angle around a point (see Example 20, page 24) 
consists of certain four angles. The second angle is twice the 
first; the third exceeds 3 times the first by 10°; the fourth 
exceeds the sum of the second and third by 10 . How large is 
each of these angles ? 




110 


ALGEBRA 


17. The second angle of a certain triangle exceeds the first 
by 30°; the third angle exceeds the second by 15°. How large 
are the angles ? 

18. The elevation of Mt. Whitney, in California, the highest 
point recorded in the United States, is 14,501 feet, measured 
from sea level. The lowest point of dry land in the United 
States is in Death Valley, California. If 52 times the elevation 
of Death Valley be diminished by 45, and the result be increased 
by the elevation of Mt. Whitney, the sum is zero. Find and 
interpret the elevation of Death Valley. 

19. There are three consecutive even integers such that the 
sum of 3 times the first, four times the second, and 3 times the 
third is 200. What are the integers ? 

20. How many pounds of 50jzf tea and how many pounds of 
90^ tea must be taken to make a mixture of 100 lb. of tea worth 
60^ per pound ? 


EXERCISE 58 

Interest Problems 

1. What is the simple interest on $500 at 6% for 1 year? 
For 3 years ? For n years ? 

2. What is the simple interest on P dollars at 6% for 1 
year ? For 4 years ? For t years ? 

3. What is the simple interest on $300 at r% for 1 year? 
For 3 years ? For t years ? 

4. What is the simple interest on P dollars at r% for 1 
year ? For 5 years ? For t years ? 

6. The total income from two investments is $135. Part 
of the money is invested at 6% and part at 7%. The total 
amount invested is $2000. Find the two sums invested. 
Solution. — 1. Let s = the number of dollars invested at 6%. 

2. 2000 — s = the number of dollars invested at 7 %. 

3. .06 s = the interest on the first sum, 



SIMPLE EQUATIONS 


111 


4 . and .07(2000 — s) = the interest on the second sum. 

5. .*. .06 s + .07(2000 - s) = 135. 

(Complete this solution and check it.) 

Note. — The essential steps may be charted as below: 



No. of Dollars 

Rate of Interest 

Annual Interest 

First sum 
Second sum 

S 

(2000 - s) 

6% 

7% 

.06 s 

.07(2000 - s) 


.*. .06 s + .07(2000 - s) = 135. 


6. One sum is invested at 6% and a second sum at 8%. 
The total income is $228. The second sum is 4 times as large 
as the first. What are the sums? 

7. A man has one sum invested at 5%, and a second sum, 
$1500 larger than the first, invested at 7%. His total income 
from these sums is $321. How much has he invested at each rate ? 

8. A man has $2500 invested at 5%. How much must he 
invest at 8% to make his total income 6% of the total amount 
invested ? 

9. If $1500 is invested at 5%, and $2000 at 6%, how much 
must be invested at 7% to make the total income 6% of the total 
amount invested ? 

10. The total of certain two investments is $2500, — one 
sum at 5%, and one at 6%. The annual interest from the 
former is $29 less than that from the latter. How much is 
invested at each rate ? 

85. Distance, rate, and time problems. 

If a train travels 40 miles per hour for 6 hours, it goes 240 
miles. 

Observe the rate (r), 40 miles per hour; the time (0, 6 hours; the 
distance (d), 240 miles. 

The distance is expressed as a number of units of length; as 
feet, rods, miles. 







112 


ALGEBRA 


The distance equals the rate multiplied by the time. (d = rt ) 

In the illustration above, 240 = 40 X 6. 

The time is expressed as a number of units of time ; as minutes, 
hours, days, etc. 

The time equals the distance divided by the rate, (t = d + r) 

In the illustration above, 6 = 240 -5- 40. 

The rate is expressed as a number of units of length in the unit 
of time. 

The rate equals the total distance divided by the time. 

0 = d -e- 0 

In the illustration above, 40 = 240 -r- 6. 

EXERCISE 59 

1. How far does a train go in 9 hours if the rate is: 

a. 25 m. per hr. ? 6. r m. per hr. ? c. (x + 2) m. per hr. ? 

2. How far does a train go in h hours if the rate is : 

a. 35 m. per hr. ? b. x m. per hr. ? c. (r — 3) m. per hr. ? 

3. Give the rule for finding the distance when the rate and 
the time are known. 

4. How long does it take an automobile to go 150 miles if 
the rate is: 

a. 20 m. per hr. ? c. 50 m. per d. ? 

b. n m. per hr. ? d. (r + 5) m. per d. ? 

6. How long does it take to go N miles if the rate is : 

а. 30 m. per hr. ? b. (r + 3) m. per hr. ? 

б. Give the rule for finding the time when the rate and the 

distance are known. 

7. At what rate is a man traveling who goes 300 miles in: 

a. 10 hr. ? b. h hr. ? c. (x + 4) hr. ? 

8. What is the rate if an object moves D miles in: 

a. 12 hr.? 6.3 d.? c. *hr.? 


SIMPLE EQUATIONS 


113 


9. Give the rule for finding the rate when the distance and 
the time are known. 

10. One man travels 25 miles per hour for x hours, and a 
second man travels 20 miles per hour for 2 hours less time. 

a. How far does the first man travel ? 

b. How long does the second man travel ? 

c. How far does the second man travel ? 

d. Suppose the sum of the distances traveled by the two men 
is 100 miles. Write the equation expressing this fact. 

e. Solve the equation, finding the time x, and check the solu¬ 
tion by finding the distances in parts a and c and comparing 
their sum with 100. 

11. A and B travel toward each other from points which are 
300 miles apart, — A at the rate of 20 miles per hour, and B at 
the rate of 25 miles per hour. In how many hours will they 
meet, if they start at the same time ? 

Solution. — 1. Since they start at the same time and travel until 
they meet, they travel the same number of hours. 

Let h = the number of hours they travel. 


Then for 

the time is 

the rate is 

the distance is 

one man 

h hours 

20 m. per hr. 

20 h m. 

the other man 

h hours 

25 m. per hr. 

25 h m. 


3. .*. 20 h + 25 h = 300. Since they started 300 m. apart and travel 
^until they meet. 

(Complete and check the solution.) 

12. Two men who are traveling in opposite directions at the 
rates of 18 and 22 miles per hour respectively, started at the 
same time from the same place. In how many hours will they 
be 250 miles apart ? 

13. A passenger and a freight train, whose rates are 30 miles 
and 18 miles per hour respectively, start toward each other at 
the same time from points which are 168 miles apart. How 
soon will they meet and how far will each have traveled ? 










114 


ALGEBRA 


14. A freight and a passenger train started towards each 
other at the same time from points which are 225 miles apart. 
If the passenger train’s rate was 30 miles per hour, and they 
meet in 5 hr., what was the rate of the freight train? 

15. A and B started towards each other at the same time from 
points which are 240 miles apart, and met in 5 hours. If A 
traveled 12 miles per hour more than B, what were their rates ?. 

16. Repeat Example 15, assuming A’s rate was 10 miles 
more per hour than B’s rate. 

17. A freight and a passenger train started in opposite direc¬ 
tions from the same place at the same time. Assume that the 
passenger train traveled twice as rapidly as the freight train. 
What were their rates if they were 300 miles apart in 5 hours ? 

18. Two automobiles started at the same time from the 
same point, in opposite directions. The first traveled 5 miles 
more per hour than the second. At the end of 8 hours they 
were 360 miles apart. What were their rates ? 

19. A and B started toward each other at the same time from 
points which are 252 miles apart. A traveled 4 miles more per 
hour than B. How fast did each travel if they met in 7 hours ? 

20. An automobile party is traveling 15 miles per hour. At 
what rate must a second party travel in order to overtake the 
first in 3 hours if it starts from the same place 1 hour after the 
first? 

Hint. — If the second overtakes the first, how do their “distances” 
compare? Which party traveled the longer time and how much? 

21. One automobile party is traveling 15 miles per hour. 
In how many hours will a second party, traveling 25 miles per 
hour, overtake the first, if the second party starts from the 
same place 2 hours after the first party left ? 

22. A left a certain point at 7 a.m., traveling 20 miles per 
hour. At 9 a.m., B started after him. How rapidly did B 
travel if he overtook A in 4 hours ? 


SIMPLE EQUATIONS 


115 


23. A left a certain point at 8 a.m. At 9 a.m., B started 
after him at the rate of 30 miles per hour, and overtook him 
in 3 hours. How fast had A traveled? 

24. A and B, traveling at 18 miles and 22 miles per hour 
respectively, started from the same place in opposite directions. 

a. A started at 6 a.m. and B at 8 a.m. How long had each 
traveled when they were finally 256 miles apart ? 

b. Solve the problem if B started at 6 a.m. and A at 8 a.m. 

25. A and B, whose rates are 20 miles and 25 miles per hour 
respectively, travel toward each other from points which are 
300 miles apart. 

a. How long and how far will each have traveled when they 
meet if A starts at 6 a.m. and B at 9 a.m. ? 

b. Solve the same problem if B starts at 6 a.m. and A at 
9 A.M. 

26. Chicago, and Madison, Wisconsin, are about 140 miles 
apart. If a train starts from Chicago for Madison at 1 p.m., 
traveling 35 miles per hour, and one starts from Madison for 
Chicago, at 3 p.m., traveling 30 miles per hour, at what time 
will they meet ? 

SUPPLEMENTARY TOPICS 

86. Further exercise in the use of formulae. 

c. Relation between Fahrenheit and Centigrade 
temperature readings. 

You are familiar with the Fahrenheit thermometer. 

Scientists and people in many foreign countries use 
the Centigrade thermometer. 

On the Fahrenheit thermometer, the freezing point is 
-f- 32° and the boiling point is + 212°; between, there are 
180 Fahrenheit degrees. On the Centigrade thermometer, 
the freezing point is 0° and the boiling point is 100° ; be¬ 
tween, there are 100 Centigrade degrees. 

100 Centigrade degrees = 180 Fahrenheit degrees 

or 1 Centigrade degree = f Fahrenheit degree. 













116 


ALGEBRA 


Example. — What Fahrenheit reading corresponds to 
+ 15° C.? 

Solution. — 1. 15 Centigrade degrees = 15(f) or 27 Fahrenheit 
degrees. 

2. + 15° C. means 15 Centigrade degrees above freezing. 

3. the corresponding Fahrenheit reading must be 27 Fahrenheit 
degrees above freezing, or 27° above 32°. This is + 59° F. 

4. /. + 15° C. = + 59° F. 

EXERCISE 60 

Find the Fahrenheit reading corresponding to : 

1. + 20° C. 2. + 35° C. 3. - 5° C. 4. - 25° C. 

5. Derivation of a formula connecting Centigrade and 
Fahrenheit readings. 

Solution. — 1. Let C = the Centigrade reading; 

and F = the corresponding Fahrenheit reading. 

2. C Centigrade degrees = (f C ) Fahrenheit degrees. 

3. But C° is counted from 0, the Centigrade freezing point, and (f C) 
must be counted from 32, the Fahrenheit freezing point. 

/. F = 32 + | C. 

Check. — 1. Let C = 0. F = 32 + f • 0, or 32. 

the freezing points correspond. 

2. Let C = 100. /. F = 32 + f (100), or 212. 

.*. the boiling points correspond. 

Note. — Memorize this formula. It is used as follows: 

Example 6. — Change 10° F. to the corresponding Centigrade 
reading. 

Solution. — 1. The formula is F = 32 + f C. 

F = +10; C = ? 

2. .*. 10 = 32 + f C. 

3. M 6 50 = 160 + 9 C. 

(Complete the solution, and tell what the result means.) 

Find 

7. F when C = + 40 

8. F when C = — 20 

9. C when F = 0 


10. C when F = + 60 

11. F when C = + 60 

12. C when F = — 20 


SIMPLE EQUATIONS 117 

13. Air liquefies when its temperature is reduced to — 182° C. 
What Fahrenheit temperature is this? 

14. Paraffine melts at + 55° C. What is the corresponding 
Fahrenheit temperature? 

15. Turpentine boils at 320° F., and alcohol at 172.4° F. 
What are the Centigrade temperatures that correspond ? 

16. The following record-breaking low temperatures have 
been produced. What are the corresponding Fahrenheit tem¬ 
peratures ? 

- 17° C., by Fahrenheit in 1714. - 262° C., by Dewar in 1898. 

- 102° C., by Faraday in 1823. - 269° C., by Onnes in 1908. 

b. Miscellaneous formulae. 

EXERCISE 61 

1. S — ^ \2 a + (n — l)dj is an important formula in 
mathematics. 

a. Find S when n = 20, a = — 5, and d = — 2. 

b. Find d when S = 165, n = 10, and a = 6. 

c. Find a when S = 174, n = 12, and d = 3. 

d. Find a when S = — 160, n = 16, and d = — 7. 

2. A = P( 1 + rt). 

а. Find t when A = $3600, P = $3000, and r = 5%. 

б. Find r when A = $3025, P = $2500, and t = 3§. 

3. K = |(a + 6 + 4m). 

o 

a. Find V when h =10, a = 12, b = 10, and m = 11. 

b. Find m when V = 126, a = 13, h = 9, and b — 15. 

4. c 2 = a 2 + b 2 — 2 ap. 

a. Find p when c = 14, b = 12, and a = 9. 

b. Find p when c = 15, b — 12, and a = 9. 

c. Find p when c = 16, b = 12, and a =9. 




VIII. SPECIAL PRODUCTS AND FACTORING 


87. In arithmetic, it is found necessary to memorize the 
multiplication table as an aid in multiplication, division, and 
factoring. 

Thus, you know now that 6*7 = 42 ; that 42 -5- 7 = 6 ; and that 
42 = 6 • 7. 

In algebra, also, certain forms of expressions occur frequently, 
which must be multiplied, divided, or factored mentally. 

88. To factor an algebraic expression is to find two or more 
expressions which will produce the given expression when they 
are multiplied together. 

Review the definitions of factor (§ 33) and common factor 

(§ 15). 

89. A number (or expression) which has no factors except 
itself and unity is called a prime number; as, 3, a, and x + y. 

A monomial is expressed in terms of its prime factors thus: 

12 a 3 f> 2 c = 2-2-3 -a-a-a-5-5-c. 

90. Squaring a monomial. 

Development. — 1. What does x 2 mean? On/) 2 ? (2 r 3 ^) 2 ? 

2. Find each of the following squares by multiplication: 

a. (2 xy ) 2 ; b. (Sa 2 b 2 ) 2 ; c. (-2 iV) 2 . 

3. Compare the exponent of each letter of the square with 
the exponent of that letter in the given monomial. 

Rule. — To square a monomial: 

Square its numerical coefficient, and multiply the result by 
each of the literal factors of the monomial, giving each letter 
twice its original exponent. 

Example. — Find (— 5 x 2 y z ) 2 . 

Solution. — (— 5 xV) 2 = 25 xV- 

118 


SPECIAL PRODUCTS AND FACTORING 119 


EXERCISE 62 

1. What sign does the square of any number have ? 

2. Learn thoroughly the squares of the integers from 1 to 20. 


Give at sight: 


3 . (a?bf 

10 . 

(+ 9 m 3 6 2 ) 2 

17 . 

(-1 «6) 2 

4 . (— a 2 6 3 ) 2 

11 . 

(-11 r 2 si 2 ) 2 

18 . 

f—V 

5 . (3 x 2 ) 2 

12 . 

(14 mx?/ 3 ) 2 

vii y/ 

6 . (a& 2 c) 2 

13 . 

C4 

<N 

T— 1 

19 . 


7 . ( — 5 to) 2 

14 . 

(- 15 x 2 yf 


V 13 / 

8 . (-4 xyf 

15 . 

d a 2 ) 2 

20 . 

/14 m\ 2 

9 . (—8 a 2 6) 2 

16 . 

( - |to») 2 

\15 a 2 / 

91. The square 

root of 

a monomial. 

If an expression can 


be resolved into two equal factors, it is said to be a perfect 
square, and one of the factors is said to be its square root. 

Thus, 4 a 2 6 6 is equal to 2 a& 3 X 2 ab z ; hence it is a perfect square 
and 2 ab 3 is its square root. 

Note. — 4 a 2 6 6 is also equal to (- 2 atf) X (- 2 ab 3 ), so that - 2 ab 3 
is also a square root. In this chapter, only the positive square root will be 
considered. It is called the principal square root. 

The following questions lead to the rule for extracting the 
square root of a perfect square monomial. 

Development. — 1. What sign does the square of any monomial 
have? 

2. When squaring a monomial, what do you do with the 
exponents of the literal factors? with the coefficient? 

3. In finding the square root, then, what should you do with 
the exponents of the literal factors? with the coefficient? 

4. Find the square root of each of the following monomials, 
and test the result by multiplication: 

b. 4zV; c. 16 fV; 


a. x 4 ; 


d. 25 x 2 y 2 z\ 


120 


ALGEBRA 


Rule. — 1. A perfect square monomial is positive, has a 
perfect square numerical coefficient, and only even numbers 
as exponents. 

2. To find its square root: find the square root of its nu¬ 
merical coefficient, and multiply the result by the literal factors 
of the monomial, giving each letter one half its original exponent. 

The symbol for extracting the square root is the radical sign, 
V; the vinculum is usually combined with it, V 

Example. — Find the square root of 25 m 4 n 6 . 

Solution. — V25 m 4 n 6 = 5 m 2 n 3 . 


EXERCISE 63 


Find the indicated square root: 


1 . 

V4x 2 

10 . V 225 c 6 

19 . 

aW 

2 . 

V 16 y 4 

11 . Vl96 a; 4 

20 . 

V*™ c 4 ^ 2 

3 . 

V<J TO 6 

12 . V 256 y 1 

21 . 

1 169 a- 4 

4 . 

V25 a 2 £> 4 

13 . V49 T 6 

' 196 j/ 2 

6 . 

V 64 r 2 s 6 

14 . Vi c 2 d 2 

22 . 

/400 a 6 

6 . 

Vsi S(P 

16 - Vim 2 


M21 5 4 

7 . 

Ji 1 

'9fe 2 

16 . \4f a 2 6 2 

23 . 

/l44 n 4 
'225 to 2 

8 . 

VlOO m 4 

17 . ^ 

24 . 

J900 c 8 

9 . 

V 169 a 2 i 2 

18 . V256 mV* 

’169 d 2 


FACTORING POLYNOMIALS 

92. It is not always possible to factor a polynomial. Those 
polynomials which can be factored are the products of certain 
special forms of number expressions. 

Note. — In this chapter, you will be taught some elementary methods 
of factoring. In Chapter XV, other methods are taught. 



























SPECIAL PRODUCTS AND FACTORING 121 


Case I 

Type Form: a(b -f c) = ab + ac 

93. Multiplying a polynomial by a monomial is taught 


in § 62. 


EXERCISE 64 


Find mentally: 

1 . 2 a(x — y + z) 

2. -3 x(x 2 + xy — y 2 ) 

3. + 4 cd(c 2 — cd + cP) 

4 . -5 m 2 (x 3 — y 3 ) 

6.-6 ab(a 2 — 2 ab + b 2 ) 


6. — § z(4 x 2 — 6 x — 8) 

7 . ran(12 m 3 — 3 m 2 n — 6 n 3 ) 

8. — a(c — d + e) 

9. f- £(r + s — t ) 

10. ^ f (a + 6 -{- c -F d) 


94. Factoring a polynomial whose terms have a common 
monomial factor. 

Example. — Factor 14 xy z — 70 x z y 2 . 

Solution. — 1. 14 and 70 can both be divided by 2, 7, and 14. 

14 is the greatest common factor of 14 and 70. 

We select it as part of the monomial factor. 

2. x is a factor of both 14 xy 3 and 70 x?y 2 . It is the highest power of 
x which is a common factor of 14 xy 3 and 70 x 3 y 2 . 

We select x as part of the monomial factor. 

3. y 3 appears in 14 xy 3 , but only y 1 appears in 70 x 3 y 2 . Then y 2 is 
the highest power of y which is a factor of both 14 xy 3 and 70 x?y 2 . 

We select y 2 as part of the monomial factor. 

4 . .*. the monomial factor is 14 xy 2 . 

5. 14 xy 3 -f- 14 xy 2 = y ; - 70 x?y 2 -r- 14 xy 2 = - 5 x 2 .. 

6. t .*. 14 xy 3 — 70 x?y 2 = 14 xy 2 {y — 5 z 2 ). 

Rule. — To factor a polynomial whose terms have a common 
monomial factor: 

1. Find the greatest common factor of its terms. 

2. Divide the polynomial by it. 

3. The factors are the common factor found in Step 1, and 
the quotient obtained in Step 2. 


122 


ALGEBRA 


EXERCISE 65 


Factor the following polynomials : 


1. 5 a + 5 b 

2. 7 t - 7 r 

3. 3 s — 12 x 

4. rax — my 

5. 3 ax 2 — 3 ay 2 

6. 4 rra 2 — 16 ry 2 

7. 7rr 2 + irrl 

8. P + Prt 

9. ^ na + J nZ 
10 . 2 7rr 2 + 2 tt rh 


11 . 3 x 2 — 6 x?/ + 3 y 2 

12. ma 2 + 6 ma + 9m 

13. to 2 — 4 te/ + 4 ty 2 

14. 3 x 2 + 3 x?/ + 3 y 2 

15. 2 r 2 x 2 — 4 r 2 x + 2 r 2 

16. 6 x 2 —■ 12 x 3 — 18 x 4 


17. an 2 — 7 an — 18 a 

18. |rP +%sB +itB 

19. jr ha jr hb ^ hm 

20. 9 m 2 x — 6 mx — 15 x 


21. 7 y 2 z + 21 ?/2 - 126 z 


22. 3 a& 2 — 15 a& + 18 a 

23. 2 m 3 n 2 + 4 ra 3 n — 126 m 3 

24. 4 ra 2 r 2 + 8 m 2 r — 60 m 2 

25. 24 m 2 a + 18 mna — 15 n 2 a 

26. 9 cV — 4 c 3 xy — 13 c z y 2 

27. 3 (?x — 132 cx — 135 x 

28. 3 x z y 2 + 6 x 2 y 2 + xy 2 + 2 ?/ 2 

29. | ax + \ am + \ az — | at 

30. ^ 7rr 2 /i + ^ 7r B 2 h + TrrRh 


31. Suppose that the polygon ABCDEF 
can be divided into six triangles, such that 
their altitudes are equal. Call the altitudes 
each a, and the bases b, c, d, e, /, and g . 

a. What is the area of A OBC ? A OPC? 

A OPF? etc. 

b. Indicate the sum of these areas. 

c. Simplify that sum by removing the monomial factor. 






SPECIAL PRODUCTS AND FACTORING 123 


d. Simplify the result by substituting p for (6 + c + d + e 


+ / + q)- 

32. Suppose that the altitude of A RX T is a, and the alti¬ 
tude of A RST is c. The base of each is b. 



a. Represent the area of each. 

b. Indicate the sum of these areas. 

c. Simplify the result by removing the 


monomial factor. 

33. The area of a circle whose radius is r is rrr 2 . 
a. What is the area of the circle of radius R ? 



b. How can you find the area of the ring be- °' 'JJjl 
tween the large and small circles ? Indicate this 

area. 

c. Simplify the result of Step b by removing the monomial 
factor. 

34. Suppose that, in the adjoining figure, the rectangles have 
equal bases of length m, and altitudes of 
length a, b , c, etc. 



a. Represent the area of each. 

b. Indicate the sum of these areas. 

c. Simplify the resulting expression by re- 


m m m m m m m 


moving the monomial factor. 


Case II 


Type Form: (a -f b) (a — b) = a 2 — b 2 

95. The product of the sum and the difference of two numbers. 

Note. — By “difference of” two numbers, we mean the first minus 


the second. 


Development. — 1. Find by multiplication as in § 63 the 


following products, and write the results as in part a : 

a. (x + 3)(* - 3) = x 2 - 9 c. (k + 10) (A; - 10) = ? 

b. (m + 7 )(m - 7) = ? d. (r + 9)(r - 9) = ? 













124 


ALGEBRA 


2. Observe the results in Step 1; try to find the following 
products mentally. Check the results by multiplication. 

a. (a + 6) (a — 6) = ? c. id + 4 ){d — 4) = ? 

b. (c + 8)(c - 8) = ? d. (y + 5 ){y - 5) = ? 

3. Find by multiplication (a + b)(a — b). 

Rule. — To find the product of the sum and the difference of 
two numbers: 

1. Square each of the numbers. 

2. Subtract the second square from the first. 

Example 1. — Find (5 a 2 + m)(5 a 2 — m). 

Solution. — (5 a 2 + m) (5 a 2 — m) = (5 a 2 ) 2 — (m) 2 = 25 a 4 — m 2 . 
Note. — When doing such multiplication, do not write (5 a 2 ) 2 — (m) 2 . 
Do that part mentally and give at once the result 25 a 4 — m 2 . 

This is called finding the result by inspection. 

Example 2. — Find mentally the product of 24 and 16. 
Solution. — 24 X 16 = (20 + 4) (20 - 4) = 400 - 16 = 384. 

(All this is to be done mentally.) 


EXERCISE 66 


Find by inspection: 

1. (x + 3)(x - 3) 

2. (m + 5)(m — 5) 

3. (i t 2 - 2)(t 2 + 2) 

4. (x 2 - 6) (a: 2 + 6) 

6. (ra 3 + 4)(m 3 — 4) 

6 . ia 2 b)ia — 2 6 ) 

7. (r 2 + 3s)(r 2 - 3 s) 

8. (3 a: - 6)(3z + 6) 

9. (7 m - 4) (7 m + 4) 

10. (5 x 2 - y){ 5 z 2 + y) 

11. (10 z + w 3 )(10 z — w z ) 


12. (4 m — .5 n)(4 m + .5 n) 

13. (6 z 2 — 7 ?/) (6 a: 2 + 7 y) 

14. (a& — 3 c)(ab + 3 c) 

15. (a 5 - 6 3 )(a 5 + 6 3 ) 

16. (x 2 + y 2 ) (x 2 - y 2 ) 

17. (a 4 + 5 4 ) (a 4 - 6 4 ) 

18. (a + b)(a — 6) (a 2 -f 6 2 ) 

19. (a: 3 -f- y 3 ) (a: 3 - 2/ 3 )(.t 6 + y 6 ) 

20. (3 xy 2 — 5 2 ?) (3 a:y 2 + 5 z 3 ) 

21. (Ja: - %y)(%x + i ?/) 

22. (Jm - i)(| m + i) 


SPECIAL PRODUCTS AND FACTORING 125 


23. (ft 2 -7)(if+ 7) 

24. (£x - t 3 t)(I* + t 3 t) 

25. (fa 2 - ib)(t*a + ib) 

26. (12 - 3)(12 + 3) 

31. 17 • 23 33. 53 • 47 

32. 41 • 39 34. 34 • 26 


27. (20 + 2) (20 - 2) 


28. (25) (15) 

29. (33) (27) 

30. (42) (38) 

35. 44 • 36 

36. 31 • 29 


See 

Example 2 
on page 124. 

37. 28 • 32 

38. 37 • 43 


39. What is the cost of 18 yards of gingham at 22£ per yard? 

40. What is the value of 26 dozen of eggs at 34^ per dozen ? 


96. Factoring the difference of two perfect squares. 
Development. — 1. What is the product of (a + 2) and 
(a — 2) ? What, then, are the factors of a 2 — 4 ? 

2. Find the factors of: 

a. a 2 — 9 b. m 2 — 16 c. k 2 — f d. 9 r 2 — 4 s 2 

3. What are the factors of x 2 — y 2< i 

Rule. — To factor the difference of two squares: 

1. Find the square roots of the two perfect square terms. 

2. One factor is the sum of the results; the other factor is 
the difference of the results. 

Example 1. — Find the factors of 25 r 4 — 16 f. 

Solution. — 1. 25 r 4 — 16 f = (5 r 2 ) 2 - (4 J 3 ) 2 . 

2. .*. 25 r 4 — 16 f = (5 r 2 + 4 * 3 )(5 r 2 - 4 f). 

Note. —'Examine the given expression carefully to make certain 
it is the difference of two squares, as in Step 1. 

However, when giving the factors, omit Step 1, and at once write 
what appears in Step 2. 

Example 2. — Find mentally the value of 13 2 — 7 2 . 

Solution. — 13 2 - 7 2 = (13 + 7) (13 - 7) 

= 20 X 6 = 120. 

Example 3. — Find mentally (x 2 — y 2 ) -f- (x — y). 

Solution. — 1. Since of — y 2 = (x — y){x + y), 

2. (x 2 - y 2 ) -5- (x - y) = (x + y). 



126 


ALGEBRA 


EXERCISE 67 
Factor the following when possible: 


1. 

x 2 - 

64 

12. 

49 — y 6 

23. 

64 x 2 y* — a 6 

2. 

y 2 - 

144 

13. 

196 - 

a 2 

24. 

81 t w - 169 m 4 

3. 

100 ■ 

-z 2 

14. 

256 c 2 

-d 2 

25. 

100 a?b 3 - c 4 

4. 

X 4 - 

y 2 

15. 

16 s 2 - 

- 50 y 2 

26. 

2 5 n 2 _ 3 6 } t 2 

T9 a T2T 0 

5. 

9 ra 2 

- 1 

16. 



27. 

225 m 2 - 256 n 5 

6. 

x 4 - 

y b 

17. 

9c 2 - 

r\d 2 

28. 

36 169 

7. 

x 2 - 

16 y 2 

18. 

■sV™ 2 

- 1 

x 2 y 2 

8. 

9. 

a 2 b 2 ■ 
36 < 6 

- 25 

- 1 

19. 

20. 

§ ~ 

144 x s 

49 

-2/ 6 

29. 

a 4 6 4 

25 49 

10. 

81 m 

2 - a 4 

21. 

9 ™10 
x 

- 1 

30. 

16 x 2 ra 2 

11. 

121 c 

i 6 - 16 b 2 

22. 

196 a 2 

- 6 4 


~7~ 25 


Find the values of the following as in Example 2, page 125, 
doing all mentally if possible. 

31. 17 2 - 8 2 34. 34 2 - 6 2 3 7. 33 2 - 27 2 

32. 14 2 - ll 2 3 5. 26 2 - 14 2 3 8. 46 2 - 40 2 

33. 22 2 - 8 2 3 6. 21 2 - 19 2 3 9. 27 2 - 23 2 

Obtain mentally the following quotients : 

40. (ra 2 — w 2 ) 4 - (m — n) 42. (x 2 — 36) 4- (x — 6) 

41. (m 2 - 16) 4- (m + 4) 43. (z 4 - 81) 4- (: z 2 + 9) 

44. (4 x 2 — y 2 ) 4- (2 x — y) 

45. (16 a 2 - 25 6 4 ) 4- (4 a - 5 b 2 ) 

46. (49 c 4 - 1) 4- (7 c 2 + 1) 

47. (100 m 4 - 49) 4- (10 m 2 - 7) 

48. (169 x 6 - 25) 4- (13 x 3 + 5) 

49. (121 x 2 - 256) 4- (11 x + 16) 

What binomial will divide each of the following? Give the 
resulting quotients. 


SPECIAL PRODUCTS AND FACTORING 127 


50. 9 x 2 — 16 y 2 52. 36 c 2 — 49 d 6 54. 196 d 2 — 256 

51. 25 r 4 - 81 53. 100 m 4 - 169 65. 400 x 4 - * 


97. Complete factoring. When a number or an expression 
is to be factored, all its prime factors should be found. Often 
factors selected at first may be factored again. 

Example 1. — 48 = 8 • 6 

= 4-2-2-3 
= 2 • 2 • 2 • 2 • 3. 


Example 2. — Factor 36 ty 8 - 36 tz s completely. 

Solution . — 1. 36 ty* - 36 Jz 8 = 36 t(y 8 - z 8 ) 

2. = 36 JO/ 4 + z 4 )0/ 4 - z 4 ) 

3. =36 JO/ 4 + zbO/ 2 + z 2 )0/ 2 “ 2 2 ) 

4. = 36 JO/ 4 + z^O/ 2 + z 2 )0/ + *0(2/ “ 2 )- 

Rule. — To find the prime factors of an expression: 

1. First remove any monomial factor which may be present. 

2. Then factor the resulting expression, when possible, re¬ 
writing all expressions which cannot be factored. 

Note 1. — Since only a few elementary cases in factoring are being 
taught in this chapter, you may at times get factors which are not 
factorable by any methods you know. “Factor completely,” there¬ 
fore, means factor as far as possible , using the methods taught in this 
chapter. 

Note 2. — You probably think that x 2 + y 1 ought to be factorable. 
It is not. Neither is x 4 + y A . You will learn more about this in 
Chapter XV. 


EXERCISE 68 

Factor as completely as you can the following: 


1. 6 m 2 — 6 n 2 

2. ac 2 — ad P 

3. 4 r 2 — 4 s A 

4. 5 am 2 — 20 an 2 

5. 16 x 2 y - 25 yz 2 


6. §x 2 -§ r 

7. 75 a 2 - 12 b 2 

8. 81 xt 2 - 49 xy 2 

9. 32 m 4 - 18 n 2 
10. 7IT 2 — 7 rs 2 


11. x 6 z - y A z 

12. 4 x 4 -4 y 4 

13. 16 aJ 4 - 16am 4 

14. x 12 - y 12 

15. 5 x 8 - 5 y 8 


128 


ALGEBRA 


Case III 


Type Form: (ax + b)(cx + d) 


98. The product of two binomials of the form (mx + p). 


Development. — 1. Below, (5 .x —4: y) is multiplied by 
(2 x + 3 y ). 

5 x — 4 y 5 :r y 

2s + Sy 


10 x 2 — 8 xy 

+ 15 xy — 12 y 2 

10 x 2 + 7 xy - 12 y 2 


2 x^+N3 y 
- 8 xy 
+ 15 xy 
+ 7 xy 


Observe the 
products 
which give 
the middle 
term. 


2. Similarly find the following products, and write the result 

= 10 s 2 + 7 xy - 12 y 2 

а . (2 z + 3)(3 s + 1) c. (4 y — 3)(2 y — 1) 

б. (3 m + 2)(2 m + 3) d. (3 y + 2)(2 y - 4) 

Notice that each of the products consists of three terms. 


3. Examine carefully the results in Step 2. Then try to get 
the first and third terms of the following products mentally. 
Possibly you can also get the middle term mentally. Check 
by long multiplication. 

a. (2r + 4) (3 r + 2) c. (2 s + 3)(2 5 + 1) 

b. (3 r + 5)(r + 2) d. (4 5 + 2)(2 5 + 3) 

4. In all the examples of Step 3, the only difficulty is that of 
getting the “ middle term.” In Step 1, this middle term, 
+ 7 xy, is the sum of — 8 xy and + 15 xy. - 8 xy is the 
product of 2 x and — 4 y ; + 15 xy is the product of 5 a; and 
3 y. These are called the cross products, because of the manner 
in which the arrows cross each other. 

Observe the location of these cross products in the equation 
below. _ 







SPECIAL PRODUCTS AND FACTORING 129 

When finding such products mentally, write only the second 
line, guided by the 

Rule. — 1. The first term of the product is the product of the 
first terms of the binomials. [5 x • 2 x = 10 x 2 ] 

2. The middle term of the product is the algebraic sum of 
the cross products. [Follow the curved lines: (— 4 y) • (2 x) 
= - 8 xy, and (5 x) • ( + 3 y) = +15 xy; (-8 xy) + (+ 15 xy) 
= + 7 xy] 

3. The third term of the product is the product of the second 
terms of the binomials. [(— 4 y) • (+ 3 y) = —12 y 2 ] 

Example. — Find (5 r — 6 s)(2 r + 3 s). 

Solution. — (5 r — 6 s) (2 r + 3 s) = 10 r 2 + 3 rs — 18 s 2 . 

Note. — Only the line above should be written. The thinking, 
which produces it, is: 

a. First term: 5 r • 2 r = 10 r 2 . 

b. Second term: — 6 s • 2 r = — 12 rs ; 5 r • 3 s = + 15 rs ; ( — 12 rs) 
+ (+ 15 rs) = + 3 rs. 

c. Third term: — 6 s • + 3 s = — 18 s 2 . 


EXERCISE 69 


Find mentally: 

1. (2x + l)(x + 2) 

2. (2x + l)(x + 3) 

3 . (3 x + l)(x + 2) 

4. (3 x + 1)0 + 3) 

6. (3 x + 1) (2 x + 3) 

6. 0 _ 4) (2 x — 1) 

7. (2 x — 3)0 — 5) 

8. (2 x — 3)(2 x — 5) 

9. (3 x — 2)0 — 6) 

10. (2 x — 3)(6 x — 1) 

11 . (3 a + 4)(2 a - 1) 


12. (4 a + 5) (2 o-l) 

13. (5 a + 6) (3 o-2) 

14. (6 b - 5) (2 6 + 1) 
16. (4 m- 7) (2 m + 3) 

16. (5 t - 8) (3 t + 2) 

17. (x ~ 7) (2 x + 9) 

18. (y + 11)0 + 6) 

19. (z - 14) (z + 10) 

20. (w + 20) (w — 5) 

21. (3 m + 2)(3 m + 2) 

22. (5 m— 3)(5 m — 3) 


130 

ALGEBRA 


23. 

(6 c 4- 

■ 5) (6 c + 5) 

42. 

(ii - 

- 6 xy) (5 + 6 xy ) 

24. 

(10 it 

1 

o 

?r- 

1 

43. 

(12 - 

- 5 mn) (4—3 mn) 

25. 

(12 m 

+ 5)(12 m - 5) 

44. 

(12 a 2 b — 2 c)(a 2 b — c ) 

26. 

(6 (lb ■ 

- 5) (4 ab + 1) 

45. 

(12 x 2 

+ 11 y){ 2 x 2 - y) 

27. 

(7 mn 

— 3)(5 mn — 2) 

46. 

(8 t 2 - 

- 5 3?) (6 ? - 3?) 

28. 

(9 r 2 - 

- 2) (8 r 2 + 3) 

47. 

(11 m 

2 - 4)(11 m 2 + 4) 

29. 

(13 s 2 

+ 6)(5 3? -2) 

48. 

(x 2 - 

6 y)(x? + 13 y) 

30. 

Ot? - 

- 8) (4 t? + 5) 

49. 

(12 a 

- 11 c)(12 a + 11 c) 

31. 

(10 n 3 

+ 3) (12 « 3 - 1) 

50. 

(15 m 

-4»)(6» + ») 

32. 

(9 c - 

2 d){ 5 c + 3 d) 

61. 

(19 a 

- 7 6) (3 o-6) 

33. 

(11 x ■ 

- 5y){2x +y) 

62. 

(9 c 3 - 

-2d) (7 c 3 + d) 

34. 

(8 a - 

11 6)(3 o + 45) 

53. 

{x 2 - 

o 

i-H 

1 

'jss 

35. 

(7 m + 9 s) (3 m — 5 s) 

54. 

(11 m‘ 

1 — 5 n) (4 m 2 4- 5 n) 

36. 

(9 t - 

4 *)(8 1 - 2 x) 

55. 

(6p 2 ■ 

- 7) (4 p 2 + 5) 

37. 

(2 ab - 

- 7 c)(3 ab + 5 c) 

56. 

(13 x 2 

+ 7 2 ) (5 x 2 — 2 z) 

38. 

(10 mri 2 + 7) (4 mn 2 — 3) 

57. 

(,x — < 

3 2/)(§z +y) 

39. 

(5 + 3 

i x) (2 + i) 

58. 

(6 a- 

- ib)(6a +ib) 

40. 

(3 y - 

2 z)(4 y — 5z) 

59. 

x ~ 

- 8)(§ a; + 9) 

41. 

(9 + 4 

:Z 2 )(10 -5?) 

60. 

(f a ~ 

-i)(12a - 15). 


99. Factoring trinomials of the form mx 2 + nx + p. The 

product of two binomials like (2 x + 3) and (3 x + 5) is a 
binomial of the form mx 2 -f nx + p. 

This means that there is a term containing the second power 
of x, one containing the first power (as a rule), and one free 
from x. 

The following discussion shows how to factor trinomials of 
the form mx 2 -f nx + p, when they are factorable. 

Development. — 1. Find the products : 

a. (2 x + 6) (3 x + 2) b. (3 x - 5) (4 x + 7) 

2. Factor 12 z 2 + 23 x + 5. 


SPECIAL PRODUCTS AND FACTORING 131 


Solution. — 1. The first terms of the binomials may be 2 x and 6 x, 
for their product is 12 x 2 . Place them in parentheses thus: 

(2* )(6x ). 

The second terms must be 1 and 5, since these are the only factors 
of 5. 

2. The second terms of the binomials are both positive since 5 and 
23 x are positive. Place the factors + 5 and + 1 in the parentheses 
and note the middle term which results. 

a. (2 x + 6) (6 x + 1); middle term, + 32 x. Incorrect. 

b. At once, interchange 5 and 1 in the parentheses. Thus: 

(2 x + 1)(6 x + 5); middle term, + 16 x. Incorrect. 

3. Step 2 shows that the factors 2 x and 6 x for 12 x 2 are incorrect. 

Try 3 x and 4 x for 12 x 2 , thus : (3 x )(4z ). 

a. (3 x + 1) (4 x + 6); middle term, + 19 x. Incorrect. 

b. (3 x + 6) (4 x + 1); middle term, -f 23 x. Correct. 

Check. — Does (3 * + 5)(4 x + 1) = 12 x 2 + 23 x + 5? Yes. 

Note. — This may seem a long process at first, but practice soon 

develops such skill that most of the trial of factors can be done mentally. 


* Factor: 


EXERCISE 70 


1. 2 x* + 7 x + 3 

2. 6 z 2 + 11x4-3 

3. 6 r 2 + 5 r + 1 

4. 2 s 2 + 7 s + 5 
6. 3 a 2 + 8 o + 5 


6. 4 x 2 + 8 x + 3 

7. Qy 2 + 11 y + 3 

8. 2 x 2 + 9 x + 10 

9. 3 2 2 + 10 z -f- 8 

10. 12 w 2 + 17 w -f- 6 

5 m 2 — 7 m + 2 


Solution. — 1. Since 2 is positive, its factors must have like signs 
and, since — 7 is negative, the cross products must both be negative. 
Therefore the factors of 2 must both be negative. 

(Complete the solution.) 


12. 

3 

m 2 

- 7m + 4 

17. 

3 v? 

-17w + 20 

13. 

4 

p 2 - 

-9^ + 5 

18. 

6 y 2 ■ 

- 19 7/ H- 3 

14. 

9 

s 2 - 

-9s + 2 

19. 

14 y 2 

- ny + 2 

15. 

6 

t 2 - 

■ lit + 4 

20. 

a? - 

10 xy + 24 y 2 

16. 

2 

w 2 • 

- 11 w + 15 

21. 

9a 2 ■ 

- 12 xy + 4 y 2 


132 


ALGEBRA 


22. Factor 15 z 2 + 14 x — 8. 

Solution. — 1. Since — 8 is negative, the factors of it must have 
unlike signs; since 14 x is positive, the larger cross product must be 
positive. 

2. For 15 x 2 , try (3 x )(5x ). For 8, try 2 and 4. 

a. (3x 2)(5x 4). The cross product, 3x-4 is the larger; 

therefore make 4 positive and 2 negative. 

Thus: (3 x — 2) (5 x -f 4). Middle term, 2 x. Incorrect. 

b. At once, try (3x 4)(5x 2). 'Now the cross product, 

4 • 5 x, is the larger. Make 4 positive and 2 negative. 

Thus: (3 x + 4) (5 x — 2). Middle term, + 14 x. Correct. 

23. Factor 24 m 2 — m — 10. 

Solution. — 1. The factors of — 10 must have unlike signs. The 
signs must be selected so that the larger cross product is negative, 
since — m is negative. 

2. Try (6 m )(4 m ). For 10, try 5 and 2. 

a. (6 m — 5) (4 m + 2). Middle term, —8 m. Incorrect. 

b. At once, try interchanging 5 and 2. 

Thus: (6 m + 2) (4 m — 5). Middle term, — 22 m. Incorrect. 

3. For 24 m 2 , try 3 m and 8 m. For 10, try 5 and %. 

a. Thus: (3m-5)(8m + 2). Middle term, — 34 m. Incorrect. 

b. Try (3 m — 2) (8 m + 5). Middle term, — m. Correct. 

Note. — In every case, first select the proper factors to give the end 

terms; then decide which cross product is the larger; then give the second 
terms their signs so that this larger cross product has the same sign as the 
middle term. 


24. 

2 x 2 + 5 x — 

3 

34. 

18 x 2 

+ 3xz -10 z 2 

25. 

6 a 2 - 

-5a — 

6 

35. 

8t 2 - 

- 26 tw + 21 w 2 

26. 

3 a 2 + 8 a — 

3 

36. 

16 ra s 

! + 18 mn — On 2 

27. 

4 6 2 - 

-5b- 

6 

37. 

x 2 - 

5 xy — 24 y 2 

28. 

5c 2 - 

- c — 4 


38. 

x 2 - 

wx — 110 w 2 

29. 

6 <P- 

f- 7 d — 

5 

39. 

12 r 2 

— 13 rs — 14 s 2 

© 

CO 

5h? + 7k- 

6 

40. 

9 a 2 - 

-30 ab + 25 b 2 

31. 

12 m‘ 

! — TO - 

- 6 

41. 

63 + 

2x - x 2 

32. 

12 f 

+ 8p ■ 

- 15 

. 42. 

14 - 

29 y - 15 y 2 

33. 

121 2 

- 13 xy + 3 y 2 

43. 

16 - 

40 x + 25 z 2 

♦ 1 


t 

SPECIAL PRODUCTS AND FACTORING 133 


44. ~'8-* 2 - 18 xy + 7 y 2 

45. 10 z 4 - 21 z 2 + 9 

46. 24 c 2 — cd — 10 d? 

47. 28 r 2 + 3 rs - 18 s 2 

48. 121 — 44 x + 4 x 2 

49. x 1 + xy — 56 y 2 
60. 18 z 2 — 39 z 4- 11 
51. 14 r 2 - 61 rt - 9 t 2 


52. 63 s & + s 3 - 20 

63. 40 y 2 + yw — 6 w 2 

64. 21 - 40 x + 16 x 2 

55. 27 d 2 + 15 ab - 50 6 2 

56. 40 z 2 - 18 a; - 9 

67. 36 m 2 - 84 mn + 49 n 2 

68. 12 r 2 - 43 rt - 20 t 2 
59. 20 m 2 + 9 mn — 20 n 2 


Case IV 

Type Form: (a + 6) 2 = a 2 + 2 + b 2 

To the Teacher. — Since this type form is a special case of that 
taught in § 98 and § 99, the customary treatment of it is unnecessary, 
in large part. “Completing the square” appears again in the chapter 
on quadratics. However, the following customary treatment of this type 
form is included for those who have special reasons for teaching it. 
( E.g ., it appears in most recommended courses of study.) 

If § 100 and § 101, which follow, are not taught, then Exercises 71 
and 72 may be used for timed tests on the methods of § 98 and § 99 re¬ 
spectively. A conservative median speed is 20 examples in 10 minutes. 

100. Squaring a binomial. 

Development. — 1. (a + b) 2 = (a + b)(a + b) 

= a 2 + 2ab + b 2 . (§98) 

2. Similarly, (a — b) 2 = (a — b)(a — b) 

= a 2 — 2 ab -f b 2 . 

These formulas are expressed by the rules which follow: 

Rule 1. — To square the sum of two numbers : 

Square the first number; add twice the product of the first 
and second numbers; add the square of the second number. 

Rule 2. — To square the difference of two numbers: 

Square the first number; subtract twice the product of the 
two numbers; add the square of the second number. 


134 


ALGEBRA 


Since a square whose side measures (a + b ) has for its area 
(a + b) (o + b) square inches, the following figure illustrates (a + b) 2 . 

(a + b) 2 = a 2 + 2 ab + b 2 . 



Example. — Find the square of 3 x 2 — 4 y z . 

Solution. — You have two ways, now, of obtaining (3 x 2 — 4 y 3 ) 2 . 

a. Using Rule 2 above: 

(3 x 2 — 4 y 3 ) 2 = (3 x 2 ) 2 - 2(3 x 2 )(4 y 3 ) + (4 y 3 ) 2 
= 9 x 4 — 24 x 2 ?/ 3 + 16 y 6 . 

b. Using the method of § 98: 

(3 x 2 - 4 y 3 ) 2 = (3 x 2 - 4 ^)(3 x 2 - 4 y 3 ) 

— 9 x 4 — 24 x 2 ?/ 3 + 16 y 6 . 


EXERCISE 71 

Square the following binomials by inspection: 


1. a+ 7 

12. 3 c + d 

23. 5 - 7x 

2. b + 8 

13. 4 x - 3 

24. 8 j/ - 3 x 

3. c - 5 

14. 5 x -f 2 y 

25. 3 cd — 10 

4 . d — 4 

15. 3 x — 5 y 

26. 5 r 2 — 9 * 

5. e + 9 

16. 2 x 2 - 7 

27. 2 x 2 - 11 ^ 2 

6. x 2 + 10 

17. 3 x 2 + 6 

28. 7 <? - 4 dm 

7. r 2 — 11 

18. 5m + 8w 

29. 11a 2 — 3 b 3 

8 . s 2 - 12 

19. 3 xy - 7 

30. 12 to 2 - 5 re 

9. ab + 20 

20. 4 w 2 - 5 2 

31. m + J 

10. 2 a + 3 

21. 6* + 9n 

32. re + £ 

11. 4 a - 6 

22. 3 - 4x 

33. x - £ 











SPECIAL PRODUCTS AND FACTORING 


135 


34 .y + i 
36. z — § 


Solution. — 1. 

2 . 


36. w 2 + f 38. b + f 

37. a - f 39. m - f 

40. Square 29 mentally. 

29 2 = (30 - l) 2 

= 900 - 60 + 1, or 841. 


Similarly square: 

41. 19 44. 18 

42. 39 46. 28 

43. 49 46. 38 


47. 21 

48. 31 

49. 22 


\ 

60. 42 

61. 51 

62. 52 


63. Problem. — Find a rule for squaring a number ending 
in 5. 


Solution. — 1. 35 = 3 • 10 + 5; 45 = 4 • 10 + 5; 55 = 5 • 10 + 5. 

2. Similarly, any number ending in 5 may be represented by 10 n + 5. 
Thus, for 95, n is 9, since 9 X 10 + 5 = 95. 

3. (10 n + 5) 2 = 100 n 2 + 100 n + 25 

= 100 n(n + 1) + 25 
or, n • (n + 1) hundreds + 25. 

Thus, the square of 95, in which n = 9, is 

9 • (9 + 1) hundreds + 25, or 9025. 


Rule. — To square a number ending in 6, drop the 6, multiply 
the balance of the number by the consecutive integer, and affix 
26 to the result. 


Example. — 85 2 = 7225. 

Explanation. — 8 X 9 = 72; affixing 25, the result is 7225. 

64. Find by this rule the squares of some numbers ending in 
5, such as 35, 45, 105, 115, etc. 

101. A perfect square trinomial is the result obtained from 
squaring a binomial. 

Thus, (3 x + 5 y) 2 = 9 x 2 + 30 xy + 25 y 2 . 

Then 9 x 2 + 30 xy + 25 y 2 is a perfect square trinomial. 

Similarly, (3 x — 5 y) 2 = 9 x 2 — 30 xy + 25 y 2 . 

Then 9 x 2 — 30 xy + 25 y 2 is a perfect square trinomial. 


136 


ALGEBRA 


Rule. — 1. A trinomial is a perfect square when two of its 
terms are perfect squares (9 x 2 and 25 y 2 ) preceded by plus 
signs; and when the remaining term (30 xy) is twice the prod¬ 
uct of the square roots of the perfect square terms, preceded by 
either a plus sign or a minus sign. 

2. To find the square root of a perfect square trinomial: 
extract the square roots of the two perfect square terms, and 
connect them by the sign of the remaining term. 

Example. — Is 4 x 2 + 9 i/ 4 — 12 xy 2 a perfect square ? 

Solution. — 1. 4 x 2 is a perfect square; its square root is 2 x. 

2. 9 y 4 is a perfect square; its square root is 3 y 2 . 

3. 12 xy 2 = 2(2 x)(3 y 2 ). 

4. Hence, 4 x 2 + 9 y 4 — 12 xy 2 is a perfect square, and its square root 
is 2 x — 3 y 2 . 

5. /. 4 x 4 + 9 y 4 - 12 xy 2 =' (2 x - 3 y 2 ) 2 

or = (2 x — 3 y 2 ) (2 x — 3 y 2 ). 


EXERCISE 72 

Supply the missing term so as to make perfect square tri¬ 
nomials of the following expressions, and then give the square 
roots. 


1 . ** + (?) +9 

2. m 2 - (?) + 25 

3. x 2 + (?) + 36 a‘ 

4. f - (?) + 4 t 2 

5. 16 x 2 + (?) + ! 


§ 164, page 249. 

Find the factors, if possible: 

11. <? - 12 cd + 36 <P 

12. x 4 + 14 x 2 )/ + 49 y 2 

13. to 2 — 24 mn + 144 n 2 


6. 9x 2 — (?) + 1 

7. 16x 2 - (?) + 9 

8. 4 m 2 — (?) + 49 re 2 

9. 25 o 2 + (?) + 16 5 2 
10. 64 m 4 - (?) + 9 c 2 

is given in 


14. 25 y 2 + 10 yz — z 2 

15. 9 x 2 — 30 xy + 25 y 2 

16. 16 r 2 - 56 rs 2 + 49 s 4 


Note. — Further practice in “completing the square” 


SPECIAL PRODUCTS AND FACTORING 


137 


17. 9 .r 2 + 49 y 2 — 42 xy 
13. 4 a 2 + 25 x* — 20 ax 

19. 81 x 2 + 4 y 2 — 36 xy 

20. 9 m 2 — 60 ran 2 + 100 n< 

21. 36 x 2 - 122 xy + 12 y 2 

22. 25 c 4 - 30 c 2 ^ + 9 d 2 

23. 49 a; 6 + 25 y 2 — 70 a: 3 ?/ 

24. 64 p 4 + 9 q 2 — 48 p 2 <? 
23. 9 m 2 + 49 n 2 + 63 ran 


26. 81 m 4 - 72 m 2 x + 16 z 2 

27. 49 z 2 + 36 y 2 - 84 xy 

28. 9 t 2 — 60 ftu 2 + 100 w i 

29. 144 p 2 — 84 pw + 49 w 2 

30. 121 x 2 — UOwx + 25 w 2 

31. 9 x 2 + 15 xy + 25 y 2 

32. 16 ft + 49 w 2 — 56 t 2 w 

33. 49 m 2 + 70 ran — 25 n 2 

34. 256 r 2 - 96 rs + 9 s 2 


Case V 

Type Form: (x + a)(x + b) = x 2 + (a + &)x + ab 

To the Teacher. — Since this type form is a special case of that 
taught in § 98 and § 99, the customary treatment is unnecessary. How¬ 
ever, the customary treatment has been included, for those who have 
special reasons for teaching it. (E.g., it is specified in some courses 
of study.) 

If § 102 and § 103 are not taught, Exercises 73 and 74 may be used 
as timed tests on the methods of § 98 and § 99 respectively. A median 
speed of 20 examples in 10 minutes is a conservative standard. 

102. The product of two binomials having a common term. 

Development. — 1. Find by actual multiplication the follow¬ 
ing products, and write the results as in part a. 

a. (x + 2)(x + 3) =£ 2 + 5z + 6 

b. (x + 5)(x + 3) = ? d. (m - 7)(m - 2) = ? 

c. (a + 5) (a + 6) = ? e. (s — 5) ($ — 8) = ? 

2. Observe carefully the results in Step 1. Try to find the 

following products mentally; check the results by multiplication. 

a. (b + 4) (b + 2) d. (y - 4)(p - 5) 

b. (c + 4)(c + 3) e. (r + 6)(r+10) 

c. (x - 3)(s - 7) /. (x - 5) (x - 11) 



138 


ALGEBRA 


Rule. — To obtain the product of two binomials having a 
common term: 

1. Square the common term. 

2. Multiply the common term by the algebraic sum of the 
second terms of the binomials. 

3. Find the product of the second terms. 

4. Add the results. 

Example. — Find ( ab + 2 ){ab — 11). 

Solution. — (ab -f- 2 )(cib — 11) = o 2 & 2 — 9 rib 22. 

Note. — Here, glance at + 2 and — 11; note that their sum is 
— 9, and that their product is — 22; write the result as above. 


EXERCISE 73 

Find the following products : 


1. 

(* 

+ 3)(* + 6) 

17. 

(* + 

10) (2 - 8) 

2. 

(* 

+ 2)(x + 5) 

18. 

(k 4- 

15) (fc - 8) 

3. 

(* 

+ 2)(x + 7) 

19. 

(m 4 

- 20) (m - 4) 

4. 

(a 

+ 4) (a + 3) 

20. 

(r- 

16)(r + 5) 

6. 

(a 

4- 5) (a + 6) 

21. 

(, 2 - 

• 7)(« 2 - 6) 

6. 

(b 

1 

CO 

1 

Oi 

22. 

(f- 

9)(<* + 7) 

7. 

(c 

- 7)(c - 2) 

23. 

(a - 

8 b) (a — 4 b) 

8. 

(d 

- 8 )(d - 5) 

24. 

(c + 

5 d)(c + 9 d) 

9. 

(e 

- 9)(e - 4) 

25. 

(x + 

9 y)(x -2y) 

10. 

(x 

- 11) (a: - 5) 

26. 

(r - 

12 s)(r + 2 s) 

11. 

(y 

+ 6)(y - 2) 

27. 

(xy + 5 )(xy - 14) 

12. 

(z 

4- 10) (z — 3) 

28. 

(m 2 - 

- 6 n)(m 2 4- 13 n) 

13. 

(w 

4- 11) (w - 5) 

29. 

(c 3 - 

■ 4 d) (c 3 — 15 d) 

14. 

(w 

- 12) (w 4- 3) 

30. 

(x 2 4 

- 7 y 2 ) {x 2 — 6 y 2 ) 

15. 

(x 

- 13) (a; + 10) 

31. 

(a - 

11 6) {a 4” 12 b) 

16. 

( y 

- 11) ( 2 / + 6) 

32. 

(r 2 - 

■ 9 s) (r 2 4- 20 s) 



SPECIAL PRODUCTS AND FACTORING 139 


33. ( rs — 14 t) (rs + 2 t) 

34. (t — 16 w)(t +6 w) 

35. (z 2 - 17) (a; 2 - 5) 

36. (;y 2 - 2)(i/ 2 + 19) 

37. (2 - 3 w)(z + 21 w) 

38. (m 2 - 13 x)(m 2 + 8 x) 

39. (t - 11 w)(t + 10 w) 

40. (z 2 - 16) (z 2 + 15) 

41. (m - 25) (m + 20) 


42. (a — 19 b)(a + 4 6) 

43. (a; + 26) (a; — 5) 

44. (c - 30) (c + 10) 

45. (ra + 20) (ra — 15) 

46. (# + ^)(# + -j) 

47. (* - i)(x - f) 

48. (r - *)(r - 2) 

49. (s + i)(s - 1) 

60. (t - i)(t + 1) 


51. Find the product 62 X 68. 


Solution. — 1. 62 X 68 = (60 + 2) (60 + 8 ) 

2 . = 60 2 + (10 • 60) + 16. 

3 . .*. 62 X 68 = 3600 + 600 + 16, or 4216. 


Note. — Learn to do all this solution mentally. 


Find the following products, mentally, if you can: 
62. 23 • 27 65. 43 • 47 58. 22 • 26 

53. 32 • 38 56. 52 • 58 59. 32 • 34 

54. 24 • 26 57. 41 • 49 60. 43 • 45 


61. 33 • 36 

62. 23 • 25 

63. 41 • 45 


103. Factoring trinomials of the form x 2 + px + q. 

Development. — 1. (x + 5) (x — 3) = x 2 + 2 x — 15. 

In obtaining this product, the algebraic sum of + 5 and - 3 is taken 
for the coefficient of x, and the product of + 5 and - 3 is taken for 
the third term. 

To factor x 2 + 2 x - 15, then, it is necessary to find two numbers 
whose product is - 15, and whose algebraic sum is + 2. 

2. Factor x 2 + 7 x + 12. 

Solution. — 1. Two numbers whose product is + 12, and whose 
sum is + 7, are + 3 and + 4. 

2. The factors are (x + 3) (x + 4) 

Check. — Does (x + 3)(x + 4) = x 2 + 7 x + 12? Yes. 





140 


ALGEBRA 


Rule. — To factor a trinomial of the form x 2 + px + q: 

1. Find two numbers whose algebraic product is q and whose 
algebraic sum is p. 

2. One factor is x + one number; the other factor is x + 
the other number. 


EXERCISE 74 


Factor the following trinomials : 

1. z 2 + 8 a; + 12 5. t 2 + 10 t + 16 

2. z 2 + 10 a; + 24 6. y 2 + 12 y + 27 

3. r 2 + 11 r + 24 7. z 2 + 14 2 + 40 

4. s 2 + 11 5 + 30 8. m 2 + 14 m + 33 

9. p 2 — 7 p + 12 


Hint. — Since 12 is positive, the second terms must have like signs; 
and, since the middle term is negative, these second terms, having like 
signs, must be negative. 

(Complete the solution.) 


10. z 2 - 7 x + 10 

11 . y 2 - 102 /+ 21 

12. z 2 -9z + 20 

13. a 2 - 12 a + 35 


14. b 2 - 9 b + 18 

15. m 2 — 9 m + 14 

16. r 2 - 12 r + 32 

17. 5 2 - 10 5 + 9 
18. a; 2 - 26 x - 192 


Solution. — 1. Since — 192 is negative, the,second terms must have 
unlike signs. Since — 26 is negative, and is the sum of the second 
terms, the term having the larger absolute value must be the negative 
one. 

2. Since 192 is a large number, systematically list its factors. 

(+ 1)(- 192); (+ 1) + (- 192) = - 191. Unsatisfactory. 

(+ 2)(- 96) ; (+ 2) + (- 96) = - 94. Unsatisfactory. 

(+ 3)(— 64) ; (+ 3) + ( — 64) = — 61. Unsatisfactory. 

(+ 4)(— 48) ; (+ 4) + (— 48) = — 44. Unsatisfactory. 

(+ 6)(— 32) ; (+ 6) + (— 32) = — 26. These are satisfactory. 

3. /. x 2 - 26 x - 192 = (x + 6)(x - 32). 

Note. — Step 2 should be done mentally. 


SPECIAL PRODUCTS AND FACTORING 141 


19. 

X 2 

+ 4* - 

- 12 

34. 

r 2 - rs - 90 s 2 

20. 

y 2 

+ 2y - 

- 15 

35. 

O 

rH 

1 

Is 

p — 24 p 2 

21. 

2 2 

+ 5z — 

14 

36. 

x 4 — 2 x 2 y 

-63 s / 2 

22. 

W 2 

— 3 w ■ 

- 18 

37. 

t 2 + 2 tw - 

■ 48 w 2 

23. 

t 2 

-3 t - 

28 

38. 

1 

Ho 

0 * 

1 

20 t 2 

24. 

m? 

1 - 4m 

-45 

39. 

c 6 +- c 3 — 56 

25. 

r 2 

- r - 42 

40. 

m 6 + 12 m 3 

- 13 

26. 

s 2 

+ 65 — 

27 

41. 

w* - 4b* • 

- 32 

27. 

t 2 

- bt - 

50 

42. 

a 2 + 11 xy 

-26 s / 2 

28. 

z 2 

+ 9z - 

36 

43. 

r 2 - 2 rs 2 - 

- 24 s 4 

29. 

a 2 

- 6 a- 

- 55 

44. 

a 4 + 4 a 2 c ■ 

- 21 c 2 

30. 

a 2 

■+■ 7 ab 

- 18 6 2 

45. 

6 4 + 3 6 2 - 

40 

31. 

c 2 

— 5 cd - 

-24 d 2 

46. 

* 

1 

• 12 s / 4 

32. 

w 2 

+ 6 wx 

— 40 x 2 

47. 

y 4 - ?/ 2 w - 

30 u / 2 

33. 

P 4 

+ 4p 2 

— 77 

48. 

w 2 — wz — 

72 z 2 


104. Mental algebra. The purpose of the study of special 
products and factoring is to enable you to do much of the alge¬ 
braic computation mentally. Avoid the use of “ scratch” 
paper. Do the multiplying and factoring mentally, and then 
write the results. 

Example. — Find the simplest form of 

3 a(2 a + 7)(a - 3) - 4 a(a + 5)(a - 6 ). 

Solution. — 1. 3 a (2 a + 7) (a — 3) — 4 a(a + 5) (a — 6) 

2 . = 3a(2a? + a — 21) — 4 a(a 2 — a - 30) 

3 . = 6 a 3 + 3 a 2 - 63 a - 4 a 3 + 4 a 2 + 120 a 

4. = 2 a 3 + 7 a 2 + 57 a. # 

Note 1. — The equality sign in line 2 means that the expression in 
line 1 equals that in line 2, etc. 

Note 2. — The binomials are multiplied together first (Step 2), and 
then their products are multiplied by the monomials in Step 3. Thus, 
(2 a + 7)(o - 3) gives 2 a 2 + a - 21; and - 4 a(a 2 - a - 30) gives 
- 4 a 3 + 4 a 2 + 120 a. 



142 


ALGEBRA 


EXERCISE 75 

Find (without use of any scratch paper): 

1. 2 m(4 m - 1 )(m + 2) 5. + 6(3 a - 2 b)(a + 3 6) 

2. - 3(5 - x )(6 + x) 6. - 2(2 r + 5 s)(2 r - 5 s) 

3. + 4(2 c — 3 d)(2 c — 3 d) 7. + 8(m — 11 n)(m + 4 n) 

4. — 5(ic — 4 y) (a; + 4 y) 8. — a(8 a — 7 m)(2 a + 3 m) 

Find all the prime factors of the following. (Review § 97.) 

9. 3 s 3 — 13 s 2 - 10 s 14. 42 t - 15 tx - 3 tx* 

10. 18 at 2 + 48 at + 32 a 15. 6 r 4 + 7 r 3 - 5 r 2 

11. 20 m 2 r — 125 n 2 r 16. 6 mx 1 — 13 mxy + 6 my 2 

12. 18 a 2 c 2 — 57 ac 2 + 45 c 2 17. 6 3? + 23 z 3 — 18 a; 4 

13. 9 x 2 n z - 16 w 3 18. 4 x 2 y - 12 xy 2 + 9 y z 

Solve the following equations: 

19. 2(3 z - l){x + 2) - 3(z - 2)(2 * + 1) = 40 

Solution. — 1. 2(3 * - l)(x + 2) - S(x - 2)(2 x + 1) = 40. 

2. 2(3 £ 2 + 5 x — 2) — 3(2 x 2 — S x — 2) = 40. 

3. + 10 x - 4 - 0 # + 9 a: + 6 = 40. 
(Complete the solution.) 

20. (4 y - S)(y + 2) - 4(y - 3)(y + 3) = 35 

21. (2-3 0(2 - St) - (St - 1)(3« + 1) = 2 

22. 5(2 r - 1)(3 r + 1) - (5 r + 2) (6 r - 3) = 2 

23. 3(2 a - 5)(a + 4) - 6(a + 8)(a - 2) = 9 

24. 7 (a - 1)(2 a + 1) - (7 a - 1)(2 a + 1) = - 2 

25. (5 * — 4)(5 x - 4) - 5(5 x + 2)(x - 4) = 6 

26. A certain number exceeds another number by 5. The 
square of the larger exceeds the square of the smaller by 95. 
What are the numbers ? 

27. Separate 25 into two parts such that the square of the 
larger exceeds the square of the smaller by 225. 


SPECIAL PRODUCTS AND FACTORING 143 


28. A certain number is 8 more than another number. The 
product of the two numbers exceeds the square of the smaller 
by 40. What are the two numbers ? 

29. There are three consecutive integers such that the product 
of the second and third exceeds the square of the first by 47. 
What are the integers ? 

30. There are three consecutive integers such that the square 
of the second, diminished by the square of the first, gives a 
remainder which is 11 less than the third. What are the integers ? 

31. Find three consecutive integers such that twice the 
square of the third exceeds the sum of the squares of the 
first and second by 43. 


EXERCISE 76 

Supplementary Miscellaneous Examples 


Multiply mentally: 

1. (4x + 5 t/) (4 x -5 y) 

2. (fa - 6)(f a + b) 

3. (a 2 — 3 b) (a 2 — 3 b) 

4. (5 r 2 - 7 s) 2 

5. (6 a 2 + 6) (3 a 2 -7 b) 

6. (x + 2)(x - 2)(x 2 + 4) 

7. (11 m — 5 n)(m + n) 

8. 3(a — x)(a — 2 x) 

9. (4 a — 3 6 2 ) (3 a + 4 b 2 ) 

10. — 2 m(m 2 — mn + n 2 ) 

11. 6 p(x 2 - l)(x 2 + 1) 

12. - 4(2 or + l)(3x - 5) 

13. (a 2 — 11 c)(a 2 + 5 c) 

14. (6 x 2 - y) 2 

16. (x + \)(x + f) 


16. (x - f)(x - f) 

17. (y + \){y - i) 

18. (f a + 1)(| a - 2) 

19. 3x(x -2)(2z + 3) 

20. (B 2 - AC)(B 2 + 5 AC) 

21. 2 a (a 2 - 3) (a 2 + 3) 

22. - 4(z +2 y)(x -2 y) 

23. (it -i)(%t+i) 

24. 91 • 89 

25. 72 • 78 

26. 68 • 72 

27. (x - l)(z + lX^ 2 + 2) 

28. (a + 3)(a - 3)(a 2 + 5) 

29. (3 n — 4 m) 2 

30. (6 p 2 — 5 q) 2 




144 


ALGEBRA 


Find the prime factors of the following. (Review § 97.) 

31. ax 2 —2 ax —35 a 

46. 21 a 2 -10 b 2 - 29 ab 

32. 3 c 2 + 4 c - 7 

47. m 2 x 2 — 7 mx — 44 

33. 6 sH - 7 st - 5 t 

48. 2 aw 2 — 3 aw —20 a 

34. 16 a 4 -80 a 2 + 100 

49. 169 m 2 — 26 mn -f n 2 

36. 3x 2 + 9x - 120 

60. w 6 + 72 - 18 w 3 

36. 5 r 2 + 16 r + 3 

61. 8 a 4 + 18 a 2 - 35 

37. 18 x 2 y 2 -3 xy 2 - 10 y 2 

62. 4 c 2 d — 8 cd — 21 d 

38. 3 m 2 — 30 m + 75 

63. 4 aY - 28 ay 2 + 49 y 2 

39. 2 n 2 — 14 n - 36 

64. z 2 + 2 zt - 63 t 2 

40. 18 z 2 + 17 * - 15 

66. 20 c 2 -30 cd - 140 d 2 

41. 15 t 2 — 6 t — 21 

56. 10 .t 2 - 13 xy - 30 y 2 

42. 4 r 2 s 2 +4 rs 2 — 15 s 2 

67. 9 a — am 2 

49 1 _ 9 A 

2S a 49 C 

68. a 2 — 36 6 2 + 5 ab 

44. 5 pr 2 + 5 pr — 10 p 

59. 9 x 2 + 3 x — 56 

46. 3 z 2 + 7 + 22 2 

60. 10 a 2 + 19 a — 15 

Solve the following equations. 

(See Example 19, page 142.) 

61. (3* - 5)(2x + 4) = {3x 

- 4) (2 x + 7) - 3 

62. (6 a + l)(3o - 4) - 2(3- 

CO 

II 

c\T 

1 

a 


63. (2 m — 5) 2 = (4 m — 3)(m + 1) 

64. (5 - 3*)(2 + s) - 3(1 - s)(l + s) = 10 

66. 2(5 t - 2)(3 t + 1) - 3(2 t + 3)(5 t - 2) = 49 

66. 3 y{y + 4) - (y + 1)(2 y - 1) = (y + 1 )(y + 9) 

67. 5(2 a - l) 2 - (4 a - 3)(4 a + 3) = 2(2 a + 7)(a - 5) 

68. (1 - 3 x) 2 - (2 x - l) 2 = 5(» + 1)(* - 7) - 21 

69. (4 m + 1)(3 m — 2) — 2(2 m — l)(m + 5) = 

(8 m — 3) (m — 2) 

70. (' 7z — 3)(7 s + 3) -8(5a + l)(a - 2) = (3z + ll) 2 


SPECIAL PRODUCTS AND FACTORING 145 


EXERCISE 77 

Problems about Areas 

1. a. What is the area of the rectangle whose base is (x — 2) 
inches and altitude is (x + 5) inches ? 

b. What is the perimeter of this rectangle ? 

2 . The base of a certain rectangle exceeds its altitude by 5 
inches. 

a. If the altitude is h in., what is the length of the base? 

b. What is the area of the rectangle ? 

c. What is the area of the square whose side equals the base 
of this rectangle ? 

d. Write the equation which expresses the fact that the area 
of the square is 65 sq. in. more than the area of this rectangle. 

e. Solve this equation and find the dimensions of the rectangle 
and of the square. 

Note. — The essential facts of this example can be charted thus: 



Base 

Altitude 

Area 

/ 

Rectangle 

Square 

h + 5 
h + 5 

h + 5 

h(h + 5) 
ih + 5) 2 


(h + 5) 2 = h(h + 5) + 65. 


3. a. Express the area of the square whose side is s inches. 

b. Express the base, altitude, and area of the rectangle whose 
base is 10 inches longer and whose altitude is 2 inches longer 
than the side of the square. 

c. Arrange the results of parts a and b in a chart as in the 
Note above. 

d. Write the equation which expresses the fact that the area 
of the rectangle is 200 sq. in. more than the area of the given 
square. 

e. Solve the equation, and find the dimensions of the square 
and the rectangle. Check the solution. 







146 


ALGEBRA 


4. The base of a certain rectangle exceeds its altitude by 
9 inches. A better shaped rectangle is formed by making the 
altitude 3 in. longer and the base 2 in. longer; but this increases 
the area by 78 sq. in. What are the dimensions of the two 
rectangles? (Represent the dimensions; form an equation, 
and solve it.) 

• 5. The base of a certain rectangle exceeds its altitude by 

8 inches. If the base and altitude are both decreased by 4 
inches, the old area exceeds the new area by 192 square inches. 
What are the dimensions of the rectangle ? 

6. The base of a certain rectangle is 7 feet more than its 
altitude. If the base be increased 5 feet and the altitude be 
decreased 3 feet, the area will remain unchanged. What are 
the dimensions of the rectangle? 

7 . The base of a certain rectangle exceeds twice its altitude 
by 3 inches. The area of a square whose side equals the base 
exceeds twice the area of the rectangle by 57 square inches. 
What are the dimensions of the rectangle? 

8. A man had plans for a house whose foundation was 12 
feet longer than it was wide. Finding that it was too expensive, 
he cut off 3 feet from the length and 2 feet from the width. The 
new foundation covered 158 square feet less than the old. What 
were the dimensions before and after making the change ? 

9. The main shaft of Washington’s Monument is square at 
the bottom and top. The side of the lower square exceeds the 
side of the upper square by 21 feet. The area of the lower square 
exceeds the area of the upper square by 1869 square feet. Find 
the dimensions of the two squares. 

10. The base of a certain rectangle exceeds its altitude by 7 
inches. A second rectangle is formed by making the altitude 
3 inches shorter and the base 10 inches longer. The new rec¬ 
tangle contains 75 square inches more than the first rectangle. 
What are the dimensions of the first rectangle ? 


SPECIAL PRODUCTS AND FACTORING 147 


11. A man’s vegetable garden is 100 feet longer than it is 
wide. A garden magazine article recommended that a 10-foot 
strip be reserved around all four sides as a drive for a team and 
wagon. He found that this would decrease the area of his lot 
available for gardening, by 9600 square feet. What are the 
dimensions of his garden ? 

105. Quadratic equations solved by factoring. 

A new kind of equation can now be solved. 

Example. — Find the number whose square exceeds the num¬ 
ber itself by 6. 

Solution. — 1. Let n = the number. 

2. Then n 2 = the square of the number. 

3. Then n 2 = n + 6. 

4. n 2 — n — 6 = 0. 

5. Factoring, (n — 3) (n + 2) = 0. 

6. The question is, what value, or values, of n makes the left side 
have the value 0? 

If n has the value which makes n — 3 = 0, then at once 
(n — 3)(n + 2) = 0, for 0 • (n -f 2) = 0. This value of n is + 3, 
for 3 - 3 = 0. 

Also, if n has the value which makes n + 2 = 0, then again 
(n — S)(n + 2) = 0, for (w — 3) • 0 = 0. This value of n is — 2, 
for -2 + 2 = 0. 

It seems, then, that + 3 and — 2 are both roots of the equation. 

Check. — Does 3 2 - 3 - 6 = 0? Does 9-9 = 0? Yes. 

Does (- 2) 2 - (- 2) - 6 = 0? Does 4 + 2 - 6 = 0? Yes. 

Also, both of these numbers satisfy the conditions of the problem. 

That is, 3 2 = 9; 9 exceeds 3 by 6. 

(— 2) 2 = 4; 4 exceeds — 2 by 6. 

Therefore “the number” is either + 3 or — 2. 

Note. — If we were seeking a number which satisfied the equation 4, 
or the conditions of the given problem, then either 3 or — 2 would do; 
but we are really seeking all the numbers satisfying equation 4 or the 
conditions of the problem; then both 3 and — 2 must be accepted as 
proper results. 



148 


ALGEBRA 


106. An equation like n 2 — n — 6 = 0 is called a quadratic 
equation or an equation of the second degree. 

Other examples are: 4 x 2 — 9 = 0 

and 3y 2 -^ = |y+ 9. 

5 3 

Notice that the equation has only one unknown; that this unknown 
does not appear in the denominator of any fraction; that it does appear 
with exponent 2; that it may or may not appear with exponent 1. 

Every quadratic equation has two roots, just as the equation 
in § 105. 

107. Solution of equations by factoring depends upon the 
following fact: 

If one of the factors of a product is zero, the product also is 
zero. 

Thus, 3X0 = 0; (- 5) X 0 = 0; 2 X 0 X (- 3) = 0. 

Moreover, if a product of two or more factors is zero, then 
one, or more, of the factors must be zero. 

Thus, if ( x — l)(x + 2) = 0, then x — 1 must = 0, or x + 2 must 
= 0, or else both x — 1 and x + 2 must = 0. 

Rule. — To solve an equation by factoring: 

1. Transpose all terms to the left member. 

2. Factor the left member completely. 

3. Set each factor equal to zero, and solve the resulting equa¬ 
tions. 

4. The roots obtained in Step 3 are the roots of the given 
equation. 

5. Check by substitution in the given equation. 

Example. — Solve the equation —— — = — • 

3 2 6 

Solution. — 1. M e 2 m 2 — 3 m = 35. 

2 . S 35 2 m 2 - 3 m - 35 = 0. 

3. Factor, (2 m + 7)(m - 5) = 0. 





SPECIAL PRODUCTS AND FACTORING 149 


4. 2m + 7 = 0ifm=— 

m — 5 = 0 if ra = +5. 

5. The roots are + 5 and — \. 

Note. — Step 4 should be done mentally and omitted. 


EXERCISE 78 

Solve by factoring the following quadratic equations: 


1. z 2 - 10 x + 21 = 0 

2. z 2 + 7 x + 10 = 0 

3. y 2 — 5 y — 24 = 0 

4. t 2 = 9 - 8 t 
6. s 2 + 7 5 = 44 

6. x* — x = 42 

7. y 2 = 18 y - 80 

8. w 2 — 36 = 5 w 

9. x 2 — 25 = 0 
10. 4 m 2 - 49 = 0 


11. 4 c 2 + 5 c — 6 = 0 

12. 5 m 2 — 6 = 13 m 

13. 2 a 2 + 7 a — 4 = 0 

14. 12 s 2 — 17 5 + 6 = 0 

15. 15 t 2 + 22 t + 8 = 0 

16. x 2 — 9 x = 0 

17. 2 y 2 + 7 y = 0 

18. 5 m 2 + 27 m = 18 

19. x 2 — ^ x — 1 =0 

20. x 2 + x = 2 


108. A literal equation is one in which some or all of the 
known numbers are represented by letters; as, 

2 x + a = bx 2 — 10. 

Example. — Find two numbers whose difference is a, and 
whose product is 6 times the square of a. 

Solution. — 1. Let x = the larger number. 

2. Then x — a = the smaller number. 

3. .*. x(x — a) = 6 a 2 . 

4. .*. x 2 — ax — 6 a 2 = 0. 

5. (x — 3 a)(x -f- 2 a) = 0. 

6. .*. x = 3 a, or x = — 2 a, the larger number. 

7. When x = 3 a, then x - a = 2 a, the smaller number. 

Check. — Does 3aX2a = 6a 2 ?< Yes. 

8. When x = -2 a, then x - a = - 3 a, the smaller number. 

Check. — Does ( - 2a)( - 3 a) = 6 a 2 ? Yes. 

(Continued on page 150.) 


150 


ALGEBRA 


Therefore there are two sets of two numbers which satisfy the condi¬ 
tions of the problem: one set, 3 a and 2 a ; the other set, —2a and 
— 3 a. 

These are really formulae for definite numbers satisfying the conditions 
of the problem. 

Thus, if a is 7, one set is 21 and 14; the other set is — 14 and — 21. 
(Read the problem, supplying 7 for a, and then check these numbers 
21 and 14; also — 21 and — 14.) 

The number a can have any numerical value. 

Example 2. — Solve the equation 3 c 2 — 2 c = 0. 

Solution. — 1. 3 c 2 — 2 c = 0. 

2. Factoring, c(3 c — 2) = 0. 

3. .\ c = 0, and c = f. 

Note. — In this example, the first factor is c. Following the third 
direction of the rule, c must = 0. 

Check. — Does 3 • 0 2 — 2 • 0 = 0? Does 0 - 0 = 0? Yes. 

Does 3(f) 2 - 2(f) = 0? Does3-|-|=0? Yes. 

P o 

3 

EXERCISE 79 


1. 

x? — 7 ax + 12 a 2 = 

0 

14. 

z 2 = 

h 2 - 

tx 

2. 

x 2 + 7 bx + 10 b 2 = 

0 



8 

4 

3. 

x 2 — 5 mx —6 m 2 = 

0 

15. 

•1 2 + 

3 rx 

- — = 0 

4. 

x 2 + 5 tx — 24 t 2 = i 

0 


2 + 

To"" 

5 

6. 

3 x? — 7 cx + 2 c 2 = 

0 

16. 

3z 2 

_ 5 ax 

+f 

6. 

2 x? = 5 bx + 3 b 2 




~ ~8~ 

8 

7. 

x 2 + 5 nx = 66 n 2 


17. 

7 x 2 

dx _ 

_2 d 2 

8. 

3x? + 13*c — 10 s 2 

= 0 


6 

~2 ~ 

3 

9. 

x? — 6 ax = 0 


18. 

* 2 _ 

17 rx 


10. 

15 z 2 = 2 p 2 — 7 px 



6 

12 

4 

11. 

o 

II 

ns 

rtH 

1 

n* 

Ci 


IQ 

2x 2 

. sx 

1 

% 

t—H 

<N 

12. 

5 x 2 — 3 cx = 0 


It/. 

5 

+ ~2~ 

10 

13. 

2 x 2 3 xw _2 w 2 _ 

= 0 

20. 

2 x 2 

. 9 mx _ m 2 


7 7 




20 

2 





SPECIAL PRODUCTS AND FACTORING 151 


109. Problems solved by quadratic equations. 

Example 1. — Find three consecutive even integers such that 
twice the square of the second, increased by the product of the 
first and the third, is 296. 

Solution. — 1. Let x = the first of these three integers. 

x + 2 = the second of these three integers, 
and x + 4 = the third of these three integers. 

2. 2(x + 2) 2 + x(x + 4) = 296. 

3. 2 x 2 + 8 x + 8 + x 2 + 4 x - 296 = 0. 

4. /. 3 x 2 + 12 x - 288 = 0. 

5. D 3 x 2 + 4 x - 96 = 0. 

6. Factoring, (x + 12) (x — 8) = 0. 

7. .*. x = — 12, or x = 8, the first of the three integers. 

8. When the first integer is — 12, the next two even integers are 
— 10 and — 8. 

Check. — Does 2(- 10) 2 + (- 12) (- 8) = 296? 

Does 200 + 96 = 296? Yes. 

9. When the first integer is 8, the next two even integers are 10 and 12. 

Check. — Does 2(10) 2 + 8 X 12 = 296? 

Does 200 + 96 = 296? Yes. 

In this example, two sets of even integers are obtained, which satisfy 
the conditions of the problem. 

Example 2. — Determine the base and altitude of the rectangle 
whose area is 140 square inches, and whose base exceeds its alti¬ 
tude by 13 inches. 

Solution. — 1. Let a = the number of inches in the altitude. 

2. a -f 13 = the number of inches in the base. 

3. a(a + 13) = the area of the rectangle. 

4. /. a (a + 13) = 140. 

5. a 2 + 13 a = 140, or a 2 + 13 a — 140 = 0. 

6. .*. {a + 20) (a — 7) = 0. 

7. .*. a = — 20, or a = 7, the number of inches in the altitude. 

8. — 20 is impossible as altitude of a rectangle. 

9. When a = 7 in., then a + 13, or 20 in. = the base. 

Check. — Does 7 X 20 = 140? Yes. 

In this example, while the quadratic has two roots, only one can hma 
real meaning for the given problem. 


152 


ALGEBRA 


EXERCISE 80 

1. The square of a certain number equals the sum of 5 times 
the given number and 6. What is the number ? 

2. The square of a certain number diminished by 8 times the 
number itself is 33. What is the number? 

3. If the square of a certain number be increased by 9 times 
the number and 25, the result is 7. What is the number ? 

4. The square of a certain number increased by the square 
of the number which is 2 more than the first gives the result 100. 
What are the numbers ? 

5. John is 4 times as old as Charles. If John’s age be added 
to the square of Charles’ age, the sum is 45. What are their 
ages? 

6. The number of members in Edward’s family is 2 more 
than the number in the family of his cousin, Fred. The square 
of the number in Fred’s family is one less than twice the number 
in Edward’s family. How many are there in each family? 

7. The sum of the squares of certain two consecutive integers 
is 145. What are the integers ? 

8. There are three consecutive integers such that the square 
of the first, increased by the square of the second, and this result, 
diminished by the square of the third, gives 96. What are the 
integers ? 

9. There are three consecutive odd integers such that the 
sum of the squares of the second and third, increased by twice 
the square of the first, equals 300. What are the integers ? 

10. Separate 22 into two parts such that the square of the 
larger increased by the square of the smaller is 260. 

11. Twice the square of a certain number equals the number 
itself. What is the number ? 

12. 9 times the square of a certain number is 25. What is 
the number? 



SPECIAL PRODUCTS AND FACTORING 153 


13. The square of the number of years in the age of Mary, 
increased by 4, equals 200. How old is Mary ? 

14. The number of boys in a certain algebra class was 3 less 
than the number of girls. One fourth of the square of the 
number of girls exceeded 4 times the number of boys by 5. 
How many girls and how many boys were there in the class ? 

15. The area of a certain small piece of ground is 240 square 
rods. Its length is 8 rods more than its width. What are its 
dimensions ? 

16. How long and wide must a lot be to contain 4500 square 
feet, and have the length 5 times the width? 

17. What are the dimensions of a triangle whose area is 54 
square inches if the base is 3 times as long as the altitude ? 

18. What must be the dimensions of a trapezoid whose area 
is 96 square inches, if the upper base and the altitude are to be 
equal, and the lower base is to be twice the upper base ? 

19. A certain lot is 25 feet wide and 130 feet long. The 
owner desires to add 1400 square feet to the lot by increasing 
its width by some amount, and its length by 5 times the same 
amount. What must that amount be ? 

20. What must be the dimensions of a lot containing 8000 
square feet, such that the length is 10 feet more than 3 times 
the width ? 

21. There are three consecutive integers such that the square 
of the first, increased by the square of the second, is 115 less 
than twice the square of the third. What are the integers ? 

22. A certain lot is 30 feet wide and 140 feet long. By 
increasing the width by a certain number of feet, and decreas¬ 
ing the length by twice the same number of feet, the area is 
increased 350 square feet. What is the number of feet ? 


154 


ALGEBRA 


SUPPLEMENTARY TOPIC 


110 . The highest common factor (H. C. F.) of two or more 
expressions is the product of all their common prime factors. 

Example 1. — Find the H. C. F. of 12 x 2 , 18 a?y 2 , and 24 x b y. 

Solution. — 1. Factor each expression. 

a. 12 x 2 = 2 • 2 • 3 • x 2 = 2 2 • 3 • x 2 . 

b. 18 xV = 2 • 3 • 3 • xV = 2 • 3 2 • x 3 */ 2 . 

c. 24 tfy = 2 • 2 • 2 • 3 • tfy = 2 3 • 3 • x 5 ?/. 

2. a. 2 is the highest power of 2 which is a common factor, because 
only one 2 is a factor of 18 x 3 y 2 . (The lowest exponent of 2 is 1.) 

b. 3 is the highest power of 3 which is a common factor, because only 
one 3 is a factor of 12 x 2 and of 24 x 5 y. (The lowest exponent of 3 is 1.) 

c. x 2 is the highest power of x which is a common factor, because 
only x 2 is a factor of 12 x 2 . (The lowest exponent of x is 2.) 

d. y is not a common factor, as it does not appear in 12 x 2 . 

3. .'. the H. C. F. is 2 • 3 • x 2 or 6 x 2 . 


Check. — 12 x 2 -s- 6 x 2 = 2 

18 xV v6x 2 = 3 xy 2 
24 &y -f- 6 x 2 = 4 x*y 


} 2, 3 xy 2 , and 4 x?y do not 
have any common factors. 


Rule. — To find the H. C. F. of two or more expressions: 

1. Find all of the prime factors of each expression. 

2. Select the factors common to all of the given expressions, 
and give each the lowest exponent it has in any of the expres¬ 
sions. 

3. Form the product of the common factors selected in 
Step 2. 

Note 1. — Step 2 of the solution above should be done mentally. 
Note 2. — The arithmetical coefficient of the H. C. F. can often 
be recognized at once, as it is the largest common divisor of the 
coefficients of the given expressions. 

Thus, in this example, 6 is easily recognized as the largest common 

divisor of 12, 18, and 24. 

\ 

Example 2. — a. Find the H. C. F. of 3 a 3 — 3 ab 2 and 
3 a 4 — 6 a?b + 3 a 2 6 2 . 


SPECIAL PRODUCTS AND FACTORING 155 


Solution. — 1. Factor each expression. 

a. 3 a 3 — 3 06 2 = 3 a(a 2 — b 2 ) = 3 a{a + 6) (a — b). 

b. 3 a 4 — 6 a 3 6 + 3 a 2 b 2 = 3 a 2 (a 2 — 2 ab + b 2 ) = 3 a 2 (a —b)(a — b). 

2. the H. C. F. = 3 • a{a — b ), or 3 a{a — b). 

Check. — 1. 3 a(a + b)(a — b) -r- 3 a{a — b) = a b. 

Because, 3a -r 3a = 1; (a — b) + (a — 6 ) = 1; 
and (o + 6 ) remains in the dividend. 

2. 3 a 2 {a — b)(a — b) -r 3 a{a — b) = a{a — b). 

Because, 3 a 2 -j- 3 a = a ; (a — 6 ) -r (a — 5) = 1. 

3. a + 5 and a{a — b) do not have any common factor. 


EXERCISE 81 


Find the highest common factor of: 

1. 18 a and 27 a 2 6. 8 x 2 y, 10 a?y 2 , and 14 Z 4 ?/ 3 

2. 12 b and 20 be 7. 6 ra 2 w 4 , 15 raw 3 , and 24 ra 3 w 2 

3. 10 z 2 and 25 zi/ 2 8. 7 rV, 21 rs 4 , and 49 rV 

4. 3 ra 2 and 7 ra 3 9. 16 c 4 d 4 , 40 c 3 d 5 , and 64 c 5 d 6 

5. 14 ab 2 and 22 a& 3 10. 24 z 2 y 3 , 48 xy 2 , and 96 x z y* 

11. (x + y)(x + y), and 3 x{x + y) 

12. 2 a(r — s)(r — s), and 4 a(r — s)(r + 

13. 8 c(c 2 + d 2 )^ + d), and 4 c 2 (c + d)(c — d) 

14. 15(a -f 2 b)(a — 3 b), and 9(a — 3 b)(a -f 3 b) 

15. (ra + w) 2 (ra — n), and (ra — w) 2 (ra + n) 

16. x 2 — y 2 , and x 2 + 2 xy + y 2 

Hint. — Read Step 1 of the Rule. Examine Example 2. 


17. x 2 — 2 x — 35, and x 2 + 3 x — 10 

18. 6 x 2 — x — 2, and 3 x 2 + 7 x — 6 

19. 2 c 2 + 3 cd — 5 d 2 , and 2 c 2 + 7 cd + 5 d 2 

20. x* + 3 x 3 — 18 x 2 , and a? — 7 x 2 + 12 x 

21. 3 x 2 — 15 xy + 18 y 2 > and 24 x 2 — 60 xy — 36 y 2 

22. y 2 - y - 12, ?/ 2 - 6 y + 8, and y 2 + 3 y - 28 

23. a 4 — 6 4 , a 4 + 2 a 2 6 2 + 6 4 , and 3 a 3 6 2 + 3 ab 3 

24. £ 2 -f 2 £!/ + 7 / 2 ,3 z 3 + 6 x 2 y + 3 x?/ 2 , and 5 x 2 y + 10 xy 2 5 y z 

25. 5 a?b + 5 a 2 b 2 - 10 a& 3 , and 10 a 4 + 30 a 3 6 + 20 a 2 6 2 


IX. FRACTIONS 


111. The expression - is called 
b 


fraction; it is read 


divided by b.” The dividend, a, is the numerator, and the 
divisor, b, is the denominator of the fraction; together, they 
are called the terms of the fraction. 

The denominator, b, must never be zero. (See § 68.) 


REDUCTION OF FRACTIONS TO LOWEST TERMS 

112. A fraction is in its lowest terms when its numerator 
and denominator do not have any common factor, except unity. 

Thus : a. -, -, x - are in their lowest terms.' 

3 b x - y 

b. —, - (« T ~j~ y) - are n0 £ thgjj. lowest terms. 

12 ab (x + y)(x - y) 


113. The first fundamental principle of fractions is used when 
reducing fractions to lowest terms. 

If both terms of be divided by 4, f is obtained. = 3, 
and also f = 3. The value of therefore, is not changed; 
only the form or appearance of it is changed. 


Rule. — If the numerator and the denominator of a fraction 
are both divided by the same number, the value of the fraction 
is not changed. 


Example 1.—Reduce 


3 aV - 3 aV 


Solution. — 1 


2 . 


6 ax 2 — 12 axy + 6 ay 2 
3 o?x 2 — 3 a 2 ?/ 2 __ 3 a?(x 2 — y 2 ) 


to lowest terms. 


6 ax 2 — 12 axy + 6 ay % 6 a{x 2 — 2 xy + 2/ s ) 

_ 3 aHx + y)(x -y) 
6 a(x — y){x — y) 


156 








FRACTIONS 


157 


3. a. Divide 3 and 6 each by 3, and write the quotients as below. 
Similarly, divide a 2 and a each by a ; {x - y) of the numerator and 
(x — y) of the denominator each by (x — y). 

la 1 

b. Then 3 a 2 x 2 - 3 aV _ $ cf(x + y)(g— y) 

* 6 ax 2 — 12 axy + 6 ay 2 0 ft(x — y)(x - y) 

21 1 

4 = a(x + y) 

2{x - y) 

Check. — Let a = 1, x = 2, y = 1. 

3 aV - 3 aV = 3 • 1 • 4 - 3 • 1 • 1 

6 ax 2 - 12 axy + 6 ay 2 6 • 1 • 4 - 12 • 1 • 2 • 1 + 6 • 1 • 1 

= 12-3 = 9 = 3 

24 - 24 + 6 6 2 


The result 


a(x + y) _ 1 • (2 + 1) _ 1 • 3 
2(x - y ) 2(2-1) 2-1 


3. 

2 


Since the two results are equal (f in both cases), the solution is prob¬ 
ably correct. 


Note 1 . — Part a , of Step 3, should be done mentally; only part b 
should be written. 


Note 2. — The process of crossing out common factors from numera¬ 
tor and denominator has long been called cancellation. The word is 
unnecessary. Remember that you are dividing numerator and de¬ 
nominator by some number and are crossing out the factors divided. 


Note 3. — The result a ^ x is the simplest form of the given 
. 2{x - y) 

fraction, and equals the given fraction for all values of a , x, and y. If, 
therefore, we want the value of the given fraction for any values of a , 

x , and y, we should substitute in , not in the given fraction, 

2{x - y) 

because the computation will be easier and the result th£ same. 

Thus, the value of the given fraction for a — 6, x = 15, and y — 5 
2 


is 5 ); that is 6 ‘ or 12. It is easier to get the result in this 

15 — 5 X0 

3 _ 0 a 2y2 

way than to substitute these values in the fraction Q ax 2_i2 axy + Q a '^’ 















158 


ALGEBRA 


Rule. — To reduce a fraction to lowest terms: 

1. Factor completely the numerator and denominator. 

2. Divide numerator and denominator by all their common 
factors, and write the quotients, including any l’s obtained as 
quotients. 

3. For the numerator of the result, take the product of the 
numerator quotients; for the denominator, the product of the 
denominator quotients. 

q w 2~2 _ q w 2,.2 

Example 2. — Reduce-r—- ——to lowest terms. 

3 m?(x - y)(x + y) 

Solution. — 1 . 3mV ~ 3 - 3 mV - f ) _ 

3 m\x - y)(x + y) 3 m\x — y)(x + y) 

111 1 

2. — 3 - 1 or i. 

3 rf{x — jr)(*-Hr) 1 

111 1 

Note 1. — Do not fail to write 1 when you get it as quotient. It 
helps to avoid a certain error often made by pupils. 

Note 2. — The result 1 means that the given fraction has the value 
1, no matter what may be the values of m, x, and y. Thus, if m = 96, 
x = 2, and y = — 37, the value of the fraction is 1. We know it 
without substituting because the fraction in its lowest terms is 1. 


114. A serious error in reducing fractions is illustrated by 
the solution below, written by a pupil. 

1 

2 a -j- b _ 2 (f, -f- b _ 2 -f- b _ • 

ac (jic c 


To prove that the result is wrong, let a = 2, b = 4, and c = 3. 
2a + 6_4 + 4_8_i 1 . 2 + 6_2+4_6 


i±i = §orli;2±fc 
6 6 3 ’ c 


or 2. 


The error occurred when a of the numerator was divided by a. 
The numerator can be divided only by factors of the numera¬ 
tor, and a is not a factor of the numerator. It is a factor only 
of one term of the numerator. 













FRACTIONS 


159 


EXERCISE 82 


Reduce to lowest terms : 

1 16 4 7 m 2 

30 16 mn 

2 — 5!^- 

‘44 ‘24 rs * 2 

3 5? 12 .tV 

‘ 72 * 34x 2 i/ 4 

10 3:2 4- 3 a; — 18 
x 2 - 8 x + 15 


? 12 a 2 6 2 c 3 (x - y) 

18 a6 2 c(x — y) 
g 26 m(x + y){x - y) 
36 ra 2 (x + y) 

(r - s)(r + 2 s) 

(r - s) 2 (r + s) 

2^ 2 ax 2 -\- 7 ax — 15 a 
4 cx 2 + 19 cx — 5 c 


11 « 2 ~ 7 a - 18 

ft 2 “h 3 -J- 2 

12 5 c 2 + 15 c - 20 

c 2 + 6 c - 7 

13 2 a 2 - 2 x - 24 

5 x 2 — 25 x — 20 
ax 2 — 3 axy + 2 ay 2 

6x 2 — 6?/ 2 

15 m2 — 7 mn + 12 n 2 
m 2 — 9 mn + 20 n 2 

16 5 x 2 — 45 ?/ 2 

4 x 2 + 4 xy — 24: y 2 

17 6 a 2 — a — 1 

6 a 2 — 5 a +. 1 

18 8 c 2 — 10 cd — 3 d 2 
12 c 2 — 5 cd — 2 d? 

19 * 4 - y 4 

‘ X 2 - 2 xy + y 2 

20. _ < ■ " -£_ 

x 4 + 2 x 2 ?/ 2 + 2Z 4 


22 12 a 2 - 60 a + 75 
8 a 2 + 36 a - 140 

2 x 2 + 12 xy - 32 y 2 

5 x 2 - 20 y 2 

OA b 2 c 2 - 121 6 2 d 2 

24. - 

c 3 + 14 c 2 ^ + 33 cd 2 

05 8 mx 2 — 2 mx — 16 m 
3 x 2 + 12 x + 12 

96 ~ 16 

x 4 — 8 x 2 + 16 

27 _ x 2 — a 2 _ 

(x + a)(x 2 + ax — 2 a 2 ) 

3 m 4 — 3 m 3 n — 18 m 2 n 2 

2 m 2 n 2 - 18 n 4 
20 x 3 - 125 xy 2 
20 x 2 y + 42 xy 2 — 20 y z 

30. _ 4 -^. ~ 4 - 

(4 x 2 + 4 xy){x - y) 

(2a — 2b) (a 2 + 4 ab + 36 2 ) 

(2 a 2 — 2 b 2 )(a + 3 6) 






























160 


ALGEBRA 




MULTIPLICATION OF FRACTIONS 


115. In arithmetic , the product of two fractions is the product 
of their numerators divided by the product of their denominators. 
Thus, f X n = 

In algebra , the same rule is followed. 

Example 1. — Find — • —— • 
t 2 x 10 a 2 


Solution. — 1. 


5 a 3 x 2 _ 15 ax 2 
2 x 10 a 2 20 a 2 x 


2. Reducing to lowest terms: 


15 ax 2 
20 a 2 x 


3 1 x 

n df __ 3_x 
20 (ft 4 a 

4 a 1 


However, it is preferable to remove any factors common to numera¬ 
tor and denominator by dividing both numerator and denominator by 
factors common to them before multiplying. 

1 x 

Thus — • ^ x2 = ^ d . 3 _ 3 x 

2x 10 a 2 2± Vbft 4 a 
2a 


Rule. — To find the product of two or more fractions: 

1. Find all of the prime factors of the numerators and de¬ 
nominators. 


2. Divide out factors common to a numerator and a de¬ 
nominator. 


3. Multiply the remaining factors of the numerators for the 
numerator of the product, and of the denominators for the de¬ 
nominator of the product. 


Example 2. — Find 


a 2 -f 3 a — 18 
a 4 - 5 a 3 + 6 a 2 


2 a 3 - 4 a 2 
ft 2 — 36 


Solution. — 1. 

a 2 + 3 a - 18 . 2 a 3 - 4 a 2 = (a + 6)(a - 3) . 2 a 2 (a - 2) 

a 4 — 5 a 3 + 6 a 2 a 2 — 36 a 2 (a 2 — 5 a 4- 6) (a — 6) (a + 6) 














FRACTIONS 


161 


2 . 


3. 

Check. — Let a = 1. 

- 2 + 3 a - 18 2 a 3 


1 1 

^(a-—^3) (it-2) 


1 , 1 

2 jr(o—2) 

(a - 6)(o-+-fi) 

1 


a — 6 


<r 


4a 2 _ 1 + 3 - 18 2-4 


a 4 -5a 3 + 6a 2 a 2 - 36 


1-5 + 6 
2 1 
-Z4 -2 


1 - 36 


-6 


1 - 6 
2 

-5 
_ 2 
5' 


+ 2 -33 
1 5 

= ±|,or-| 

— 5 5 

Note. — The final result equals the indicated product for all values 
of a, except any which make a denominator zero. For example, a can¬ 
not be 6 in this example, because when a = 6, a 2 — 36 is zero and 
the second fraction does not have any meaning then. 


EXERCISE 83 

Find the following indicated products : 


16. 


17. 


1 . 

5 


14 

6 . 

3 

14 11 


11 . 

2 xy 2 . 

3 xy 


6 


5 


10 

33 7 



3 x 3 

y 2 

2 . 

3 


20 

7. 

4 . 15 . 7 


12 . 

4 a 3 6 _ 

15 x 2 y 


8 


15 


5 * 14 ’ 6 



5 a : 2 

4 a?b 

Q 

5 


6 

ft 

6 

14 # 10 



5 a 2 

6 be 2 

O . 

2 


33 

O • 

21 ’ 

5 * 4 


1 O . 

4 be 10 a 2 bc 

4. 

3 

a 


9. 

2 a 

6 6 5 

c 

14. 

27 xV 

30 xz 2 


4 

6 

5 a 

S~b 

’ 7c ’ 4 

a 


20 £ 4 £ 

18 y 3 

5. 

6 

X 

. 10 y 

10 . 

2 a 6 

. 3 6 c # 

5 ac 

15. 

10 m 2 n 

3 . 3 nV 


5 

y- 

2 3 a: 2 

3c* 

* 4 a 2 * 

"662 

6 nr 3 

5 m A n“ 


x + y ' 3 x 2 

2 x (x + y) 2 
a 2 -b 2 5 a 3 

40 s 


(a + bf 


18. 


19. 


6 m + 12 5 m — 15 


3m - 9 
am + ax 


4m + 8 
cm — cx 


dm — dx bm + bx 































162 


ALGEBRA 


a 2 - 6 2 

ac -f 6c 

22 . * - 16 , • 

z 2 - 7 X + 10 

(a + 6) 2 

ac — 6c 

(* - 2)* 

z 2 — 9 x + 20 

2 t + 8 

5* — 10 

03 3 y — 15 

. y 1 + 9 y + 8 

< 2 -4 

t 2 + 6* + 8 

2 y + 8 

2/ 2 + 3 y - 40 


24 . * 2 - y 2 . 4 X - 8 y 

x 2 — 4 y 2 3 z 2 + 3 xy — 6 y 2 

2 ^ a 2 x — a 3 x 2 — ax — 6 a 2 

v? — 3 az 2 x 2 + aa: - 2 a 2 

26 q2 ~ 6 — 27 # a 2 + 10 a + 24 

' a 2 + 9 a + 18 * a 2 - 7 a - 18 

2? 2 r 2 + 5 r<? - 12 s 2 r 2 - 2 rs + s 2 

2 r 2 — 5 rs + 3 s 2 r 2 + 5 rs + 4 s 2 

28 ^ a ~ 6 6 ' 3 a 2 + 2 ab — b 2 a 2 — ab —6 b 2 

’9 a 2 —b 2 a 2 — 4 b 2 6 a + 6 6 

29 a4 ~ b 4 . 4 a 2 - 6 2 # 6 a 2 + 6 6 2 

3a—36 a 4 + 2 a 2 6 2 + 6 4 2 a 2 -f a6 — 6 2 

4 z 2 + 12 x + 9 . 3 a: 2 + 11 x - 20 3 z 2 + 22 z + 7 

' 3 z 2 + 17 x - 28 ’ 2 z 2 - x - 6 ’ 2 z 2 + 13 x + 15 

2 ^ 3 m 2 — m < 2 m 2 +15 m -f 25 # 6 m 2 - 15 m 

4 m 2 — 25 6 m 3 m 2 + 14 m — 5 

32 4 ~ x<2 . 20 — x - x 2 > 6 - 2 a: 

' 12 - 3 x 4—4z + z 2 10 + 7z + x 2 

33 a x 2 — mo; + m 2 > 2 x 2 - 2 mx > x 2 — 2mx - 3 m 2 

2 x 2 — 2 m 2 5 x 1 — 15 mx 3 a: 2 — 3 mx + 3 m 2 
b. What is the value of the indicated product in part a when 
x = 3 and m = 2? 

4 x 2 — 25 2 / 2 b 15 x 2 — 55 xy — 20?/ 2 2x — 4y 
10 a: - 20 y' 2 x 2 - 3 xy - 20 y 2 * 6 z 2 - 13 xy - 5 ?/ 2 







































FRACTIONS 


163 


DIVISION OF FRACTIONS 


116 . In arithmetic , to divide one fraction by another, we 
multiply the dividend by the inverted divisor. 


Thus, a. 


3 

JL§ = JL x 0 = 3 

10 ‘ 5 Z0 3 2* 

2 1 


b. 3 j + 2i = n 

In algebra, we do the same. 


n = n ^ = 5 

5 3 XX 3* 

1 


Rule. — To divide one fraction by another: 

1. Factor the numerators and denominators of the fractions. 

2. Invert the divisor fraction. 

3. Multiply the dividend fraction by the inverted divisor. 
Note. — The divisor follows the sign of division. 


Example. — Divide 


a 2 m + 10 am + 21 m ^ a 2 m 3 


9 m 3 


4 a 2 + 3 a 


a° — or 


Solution. — 1. 


2 . 


3. 


4. 


q 2 m + 10 am +21 m q 2 m 3 — 9 m 3 
a 3 — 4 a 2 + 3 a * a 3 — a 2 
_ m(a 2 + 10 q + 21) m 3 (a 2 — 9) 

a(a 2 — 4 a + 3) * a 2 (a — 1) 

1 1 a 1 

_ -m(q + 7)(a-+-3) _ -q 2 ^-1) 

-q(q — 3)(q—"4) -^(q — 3)(u-+-3) 

1 1 m 2 1 

- q(q + 7) ^ 
m 2 (q — 3) 2 


CTiecfc. — Let a = 2; m = 1. 

q 2 m + 10 qm + 21 m q 2 m 3 — 9 m 3 _ 4*1 + 10 • 2 • 1 + 21 4 — 9 

q 3 — 4 q 2 + 3 q q 3 — q 2 8 — 16 + 6 8 — 4 

1 1 

q(q + 7) _ 2(2 + 7) = 2-9 _ 18 _ 1R 
m 2 (q — 3) 2 1(2 — 3) 2 1(— l) 2 1 


Also, 


















164 


ALGEBRA 


EXERCISE 84 

3_ ^2 3 18 x 2 3 x 36 m 2 . 9 m 3 

14 : 7 5 ?/ 2 ’ 10 y 35 rc 3 * 14 a 2 

9^^15a a 12 c 2 rf . 4c? 6. 24zV-r-^^ 

4 6*86 * 28 erf 2 * 7 c 4 

5 m _._ 10 m 

' x 2 -8x + 15 ’ x 2 -2x - 15 
2 c ^ 4 c 2 1Q c 2 + rf 2 c 4 - rf 4 

4c 2 — 1 6 c + 3 * c + rf c 2 — 2 erf H- rf 2 

5a+ 15 5a-15 n 1 2 - 4a; . 9 - 6 a + a 2 

* a 2 - 9 a 2 — 6 a + 9 * 10 + 5 x ' 6 + 5 x + x 2 

12 3,2 ~ X V : x<2 ~ V 2 

3 x — 6 y x 2 — xy — 2 y 2 

13 Q 3 + a6 2 a 2 + 6 2 

a 3 — 2 a6 2 a 2 — 4 a6 + 4 6 2 

14 (2 r + g) 2 4 r 3 — rs 2 

8 r — 4^ 8 rs — 4 s 2 

15 # 2 — £ ~ 42 x 2 + 2x — 8 
x? + 5 z - 84 * x 2 + 10 # — 24 

m 3 —5 m 2 — 6m 2 m 2 — 14 m + 12 
3m 2 + 9m + 6 m 2 + m — 2 

c 2 — 5 erf + 6 rf 2 . c 2 + 2 erf — 15 rf 2 

7 * c 2 - 9 erf + 14 rf 2 * c 2 - 2 erf - 35 rf 2 

12 a 2 - 17 a6 - 5 6 2 . 8 a 2 - 2 ab - 6 2 

9 a 2 - 30 a6 + 25 6 2 * 9 a 2 - 18 a6 + 5 62 

a 2 - 16 . a 2 - 16 a + 55 . 3 a - 12 

a 2 — 9a — 5 a 2 — 7a — 44 4 a 2 — 1 

2 g (2 # — y) 2 < 4 xy — 12 y 2 2 >t 2 — 7 xy + 3 y 2 

' 3 x* + 15 xy 12 z 2 — 4 xy — y 2 6 x 2 + 31 xy + 5 y 2 

/ c 2 + rf 2 e 4 + 2 e 2 rf 2 + rf 4 \ # e 3 + erf 2 
ve 2 — rf 2 c 2 + 2 erf + rf 2 / erf + rf 2 



































FRACTIONS 


165 


ADDITION AND SUBTRACTION OF FRACTIONS 

117 . In arithmetic, fractions having a common denominator 
are added or subtracted without difficulty: 

Thus, a. A + A = H- A ~~ A = A- 

Fractions which do not have the same denominator must be 
changed to equal fractions which have a common denominator. 

Thus, l+i + i = A + A+ A or H- 

12 , the lowest common denominator, is the smallest number which 
contains 3, 2, and 4 exactly as divisors. 

In algebra, also, before fractions can be added or subtracted, 
they must be changed to equal fractions having a common 
denominator. 

You must, therefore, learn: 

a. how to form the lowest common denominator; 

b. how to change fractions to “ equal ” fractions having this 
common denominator. 

118 . Lowest common multiple. 

a. A multiple of a number is any number which contains the 
given number as an exact divisor. 

Thus, 6, 9, 12, and 30 are multiples of 3. 

b. A common multiple of two or more numbers is a multiple 
of each of them; it can be divided by each of them. 

Thus, 24 is a common multiple of 2, 3, 8, and 12. 

Also, 48; 72, 96, 120, etc., are common multiples of 2, 3, 8, and 12. 

c. The lowest common multiple of two or more numbers is 
the smallest of their common multiples. 

Thus, 24 is the lowest common multiple of 2, 3, 8, and 12. 

d. The lowest common multiple of two or more expressions 
is that multiple of them which has the least number of prime 

factors. 


166 


ALGEBRA 


Example. — What is the lowest common multiple of 5 a 3 6 
and 2 ab b c ? 

Solution. — 1. The smallest number which will contain both 5 and 
2 is 10, — namely, 5 • 2. 

2. The lowest power of a which will contain o 3 and a (separately) is 
o 3 ; that is, a 3 -5- a 3 = 1; a 3 -s- a = a 2 . 

3. The lowest power of b which will contain b and b b is b b . 

4. The lowest power of c which will contain c is c. 

5. the L. C. M. = 10 a*b b c. 

Check. — Does 10 a?b b c contain each of the expressions? 

Yes, for 10 a l , [' e = 2 b‘c ; and X ° ^ = 5 a 2 . 

5 a?b 2 ab 5 c 

Rule. — To find the L. C. M. of two or more expressions: 

1. Find the prime factors of each of the expressions. 

2. Select all of the different prime factors and give to each 
the highest exponent with which it appears in the given expres¬ 
sions. 

3. Form the product of all of the factors selected in Step 2. 


Example. — Find the L. C. M. of 2 am?(m + a) and 
9 a 2 (m + a) 3 . 

Solution. — 1. 2 flm 2 (m + a) = 2 • a • ra 2 • (ra + a). 

9 a 2 (ra + a ) 3 = 3 2 • a 2 • (m + a) 3 . 

2 . 2 appears with 1 as its highest exponent. 

3 appears with 2 as its highest exponent. 
a appears with 2 as its highest exponent, 
m appears with 2 as its highest exponent. 

(m + a) appears with 3 as its highest exponent. 

/. L. C. M. = 2 • 3 2 • o 2 • m 2 • (m + a ) 3 
= 18 aV(m + a) 3 . 


Check. — Does 18 a 2 m 2 (m + a) 3 contain each of the given expressions? 


Yes, for 


18 o 2 m 2 (m + o) 3 _ 


= 9 a{m -f- a) 2 


2 am J (m + a) 

18 a 2 m 2 (m + a ) 3 —2 m 1 
9 a 2 (m + a) 3 


and 






FRACTIONS 


167 


EXERCISE 85 

Find the L. C. M. of the numbers in each example, and 
check: 


1. 6 and 8 

2. 12 and 15 

3. 18 and 12 

4. 6 a and 9 b 

5. 4 x 2 and 10 y 2 

6. 4 ab and 6 a 2 b 

7. 8 mn and 12 n 2 


8. 4 r 2 and 14 r 3 

9. 16 r 4 and 12 rs 3 

10. 30 m 3 and 18 mn 4 

11. 5 a 2 , 10 b 2 , and 15 c 2 

12. 4 c 3 , 6 c 2 d, and 10 cd? 

13. 20 x, 15 xy, and 12 x 2 y 

14. 12 y z , 8 x 2 y, and 15 x 3 


15. 27 c 4 , 6 c 3 d, and 18 c 2 <f 

16. (x + 2)(x + 5) and (x + 2)(x — 3) 

17. (m — 2 n) 2 and (m + n) 

18. (1 -f a) 3 , (1 — a)(l + a), and (1 -f a) 2 

19. 3 (a + b) 2 , 4(a — b)(a + b), and 2(a — 6) 2 

20. 2 Z - 10 and 3 x — 12 

21. 3 a — 7 and 6 a + 15 

22. 5 a + 5 b and a 2 — b 2 

23. 4 p 2 — 1 and 2 p + 1 

24. x 2 — y 2 and x 2 — 2 xy + y 2 

25. mx — my, nx + ny, and x 2 — y 2 

26. x 2 — 9 and x 2 — 8 x — 33 

27. 4 c 2 — 25, 6 c — 15, and 8 c -f 20 

28. a 2 + 2 a& + b 2 and 2 a 2 + ab — b 2 

29. 4 a 2 — 9 6 2 and 4 a 2 -f 12 a& + 9 b 2 

30. 9 mx 2 — m, 3 mnx — mn, and 9 mr 2 — 6 nx + n 

31. x 2 — 10 x -f 21, x 2 — 3 x — 28, and z 2 + x — 12 

32. 6 r 2 + 11 rs — 10 s 2 and 12 r 2 + rs — 6 s 2 

33. 10 a; 2 - x - 21, 15 x 2 + x - 28, and 6 x 2 - 17 x + 12 


168 


ALGEBRA 


119. The second fundamental principle of fractions. 

If both terms of the fraction % are multiplied by 4, the result 
is iy*. Since both fractions have the value 3, they are equal. 

Rule. — If the numerator and denominator of a fraction are 
both multiplied by the same number, the value of the fraction 
is not changed. 

This principle is used in changing fractions to a common 
denominator. 

120. The lowest common denominator (L. C. D.) of two or 
more fractions is the lowest common multiple of their de¬ 
nominators. 

Example 1. — Change to their lowest common denominator 
the fractions f and 

Solution. — 1. The L. C. D. of these fractions is 12, for 12 is the 
L. C. M. of 4 and 6. (§118.) 

2. To change the denominator 4 into 12, it must be multiplied by 3; 
that is, by 12 4 - 4. Then, in order not to change the value of the 
fraction, multiply also the numerator by 3. 

• ? = 3j_3 = 9_. 

4 3-4 12* 

3. For the fraction f: 12 6 = 2. Multiply numerator and de¬ 
nominator by 2. 5 _ 2 • 5 _ 10 

6 ~ 2^6 ~~ 12* 

The fractions have been changed to ^ and If. These fractions 
have the common denominator 12. They have the same value as the 
fractions f and f, because they were derived by use of the rule in § 119. 

In algebra, use the following 

Rule. — To change fractions to their lowest common de¬ 
nominator : 

1. Find the prime factors of the denominators. 

2. Find the L. C. M. of the given denominators; this is the 
L. C. D. 

3. For each fraction, divide the L. C. D. by the given denomina¬ 
tor ; multiply both numerator and denominator by the quotient. 


FRACTIONS 


169 


Example 2. — Change to their lowest common denominator: 
4 a , 3 a 

-o- 7 and "2 - r - 

a z — 4 a - 5 a + 6 


Solution. 

1 4 q _ 

4 a 

a 2 - 4 

(a - 2) (a + 2) 

2. 

3 a 

3 a 

a 2 - 5 a + 6 

(a - 2) (a - 3) 

3. 

The L. C. D. is 

(a + 2) (a — 2) (a 

4. 

L. C. D. 4- (q 

- 2) (a + 2) = a 


. 4 a _ 4 a(a — 3) _ 

" (a - 2)(q +2) (a - 2)(a + 2)(a - 3)’ 

5. L. C. D. 4- (a — 2)(a - 3) = (a + 2). 

. 3 a _ _ 3 a(a -j~ 2) _ 

** (a - 2) (a - 3) .{a- 2 )(a + 2) (a - 3) 


Check. — The final fractions in Steps 4 and 5 may be changed into 
the original fractions by dividing out common factors. 

Note 1. — In Step 4, to get L. C. D. 4- (d — 2) (a + 2), write, if 
necessary: 

Co—I—(a-—-2) (a — 3) _ a g 
-2)(a-+-2) 

Similarly, in Step 5: 

(q +2 )Ca-2-Ka-—3) = a + 2 

(a-«)(a^— 

Try to do this mentally, however. 


Note 2. — In Step 4, the fractions 


4 q 


(q — 2)(q + 2) 


and 


4 q(q — 3) 


(a - 2) (a + 2) (a - 3) 


are 


'equal.’ 


Really they are identically equal. (See § 76.) Such fractions are 
called equivalent fractions , to distinguish them from fractions (which 
will be studied soon) which are equal for some few values of their letters. 

Thus - = *L±_I when x = 2 but not otherwise. 

’ 2 3 

Remember: equivalent fractions always have the same value but 

differ in form. 

















170 


ALGEBRA 


EXERCISE 86 


Change to equivalent fractions having their L. C. D.: 

4 x 2 . 3 y 2 


1 

8 * 3 

2 3 - 3 

2 - 5’ 4 

Q Cl d 

3 - 6 ’ 5 

. vi n 

4 - 8 ; 5 

. 2a 3 b 

B - T ; T 


6 . 


7 . 4 

x y 

8. I; - 8 

m n 

9 . 2 . * 

X 2 ’ X 

“■ 3=5 


16 . 


2 . 

4. 6 


a 

6 ’ c 


2 a 

3 6 

4 c 

m 

’ m 2 ’ 

m 3 

a. 

6 c 


> 

a: 

a: 2 ’ x 3 


5 

3 . 

4 

2x 

> A > 
4 y 

3 2 

2 a 

. 36. 

5 J 

4 c 


m?z Tir mr 


16 . 

17 . 

18 . 

19 . 

20 . 


£ 


y 


X + y X - y 
a b 


2a — b’ 2 a + 6 
2 m 3 n 


3 m + n ’ 2 m — n 

_3_. 4 

2 r - 6 ’ 3 r - 9 
a 6 


5 a: + 10’ 7 x — 21 
26 . 3 * 

27 . 

28 . 

29 . 

30 . 


21 . 

22 . 

23 . 

24 . 

26 . 

5 x 


5 


a: 2 - 9 ’ a: + 3 
3 2 


4m 2 — 1 ’ 6m + 3 
3 r 2 r + s 
r 2 — 25 s 2 ’ r - 5s 
x* — y 2 x + 2 y 


(2 x — 3 y) 2 ’ 10 x — 15 y 

m . _ n 

(2 m -3w) 2 ’ 4 m 2 — 9 n 2 


x 2 

- 4a: - 

21 ’ z 2 

- 2 a: - 35 



3 a 


2a 

4 a 

6 a 2 — a — 

1’ 3 a 2 

! + 7 a + 2 ’ 

2 a 2 + 3 a - 2 


m + 1 


m — 4 

m + 3 

m 2 

— m — 

6 ’ m 2 ■ 

- 4 m + 3 * 

m 2 + m - 2 


r 


a 

ar 

r 2 

— 6 ar + 9 a 2 ’ i 

r 2 -f 4 ar — 21 a 2 ’ r 2 + 7 ar 


2 

1 

3 



3 x — 6y’ 2 x — 3y’ S x — 4 y 




































FRACTIONS 


171 


121. Adding and subtracting fractions is now possible. 

(See § 117.) 

Rule. — To add or subtract fractions : 

1. Reduce them, if necessary, to equivalent fractions having 
their lowest common denominator. 

2. For the numerator of the result, combine the numerators 

of the resulting fractions, in parentheses, preceding each by 
the sign of its fraction. % 

3. For the denominator of the result, write the L. C. D. 

4. Simplify the numerator by removing parentheses and 


(Rule, Step 1) 
(Rule, Steps 2,3) 
(Rule, Step 4) 
(Rule, Step 4) 
(Rule, Step 5) 


Check. — 

Let s = 1 ; 

y = 

1. 




5 x — 4 y 

II 

o* 

1 

o 

i. 1 





6 

6 

6’ 

► and — — 

5 _ 

7 

i 

loi 

II 

ji 

O 

*-t 

7s - 2 y 

7—2 
— or 

5 . 

6 

14 

42 

42 42 21 

14 

14 

14’ 





Also, 7x 

i> 1 

II 

>5 

H 

rH 

1 

- 11 _ _ 4 





21 

21 

21 





combining like terms. 

6. Reduce the result to lowest terms. 


j Example. — Find 
Solution. — 1. 

2 . 

3. 

4. 

5. 

6 . 

7. 


-Pin A 5* -4;/ _ 7* - 2y. 

6 14 

5 x — 4t/ _ 7 x — 2y 
6 14 

_ 7(5 x - 4 y) 3(7 x - 2 y) 
42 42 

= 7(5 x — 4 y) - 3(7 x - 2 y) 

42 

_ 35 x — 28 y — 21 x + 6 y 
42 

_ 14 x — 22 y 

42 

2(7 x - 11 y) 

42 

7s — 11 y . 

Ol 





















172 


ALGEBRA 


Note 1. — Use this form of solution. The equality sign in line 2 
means that the fractions there are equivalent to the ones above in 
line 1 ; etc. 

Note 2. — After a few examples, omit Step 2; pass at once from 
Step 1 to Step 3. Always write Step 3. Notice the words “in paren¬ 
theses” in the second step of the rule. 

Note 3. — Remember that the result, 


21 


is identically equal 


5s — iy_ 7_x — 2y, 
6 14 


Therefore, if we want the value of —- — 

6 


— ~ - X 2 y for any values of x and y, we should substitute in ^ x 11 y 

because that is simpler than substituting in 5 x — 4 y _ 7_ x — 2 y . 

6 14 

Thus, when x = 2 and y — 1 , the given expression in line 1 equals 
7-2 — 11-1 _ 3 1 

21 ' 21* 7 * 

EXERCISE 87 


tM 

4 6 

2 . 5-7 

5 15 

3.2-2 

8 3 

< 5 fl I o 

T + 3 


8 . 5-3 + 5 

3 4 6 

9 . - —— +1 

5 15 3 

10 . 2 £_? + 5 £ 

3 2 6 

11 -^1 + L 

'7 14 ~2 


_ 5m , 3m ._ 3 w . 3 w 3 w 

0 .-- 12 .-- 

5 5 4 2 


6 

6 3a 2a; 

2 3 

7 13 y 5 y 


15 


6 


13. ® + £ 4 . £ 

3 4 2 

14 . - — — + - 

3 5 6 


22 . 


2 a: + 5 


jg 4 o — 3 | 2 o -f- 1 


16 o + 3 3o-2 

3 5 

17 x ~ 3 y _1_ 2 g + 3?/ 

4 14 
lg 5c + 4 c + 3 

19. 

20 . 

21 . 

6 

5 x + 2 


12 

9 

2c - 1 

c —2 

5 

6 

3m + 7 

2m - 1 

5 

4 

1 

00 

5z — 4 y 


4 


6 




























FRACTIONS 


173 


23. 


24. 


25. 


3 a -f- 5 

4 

4 x + 1 

3 

6 c — 4 d 


5 o + 3 + n — 6 


6 

6 x — 1 _ 

’ <r~ 

5 c — d 


8 

4x + 3 
6 

,4c -j- 3 d 


2 ^ 3m + 2n_|_5m-2n 


5 

4 m + n 


27. 


4 9 

2 r — s 3 r -f 2 s 


3 

4r + 3^ 


10 


28. Find 
Solution . — 1. 
2 . 


2 a + b 2 6 — _c 


ab 


be 

,26- 


15 

_ 2 c + a 
ac 

c _ 2 c + a 


5 r — 4 s 
20 


2 a + 5 

a6 6c ac 

_ c(2 a + 6) a(2 6 — c) _ 6(2 c + a) . 

abc abc abc 


(Complete the solution and check it.) 

Note. — The L. C. D., abc , is gotten by the rule on page 168. 

^ a is equivalent to c ^ 2 a ^~- - • 
ab abc 

2 a + 6 _ c(2 a -f 6) 


For. abc -r a6 = c; . - , 

ab 

Similarly for the other fractions. 


abc 


(Rule, page 168) 


29. 

— + — 

5a 3a 

34. 

c a 

x y 

39. 

30. 

_7_4_ 

2 m 5 m 

35. 

2 _ 3_ 

5 x 4 y 

40. 

31. 

- + f 

a b 

36. 

a , b 

3s 7* 

41. 

32. 

6 _ 7 
r 5 

37. 

m n 

rx ry 

42. 

33. 

a 5 

x £ 

38. 

5 2 

X 2 X 3 

43. 


- 1+1 
2 a6 6 2 


2 
a 5 

1 

2x 2 
2x - 1 
3x 


J_+_5_ 

3 xy 6 ?/ 2 


+ 


3 x 4~ 2 
4 x 


5x + 3 y _ 2 x + 4 y 
x 2 y xy 2 

x 4- 1 


2 x 


x 2 - 1 
3 x 2 





































174 


ALGEBRA 


44. Find 

Solution. — 1. 
2 . 


4 p. — 6 3 p — 12 

3 4 


4 p — 
3 


3 p — 12 
4 


( Factoring the\ 
denominators / 

3 = 9(p — 4) _ 8(p — 3) /Step 1, ruleA 

6(p -3)(p - 4) 6(p - 3)(p - 4) \page 171 / 

(Complete the solution, following Steps 2-5 of the rule.) 

Note. — In Step 3, to get - 9(p — 4) - 

6(p — 3)(p — 4) 

a. 6 (p - 3)(p - 4) -b 2(p - 3) = 3(p - 4). (See Step 3, rule, 
page 168.) 

6. Multiplying both terms of ^ ^ ^ by 3(p — 4), 

we get - Q(P ~ 4) - 

6(p - 3)(p - 4) 


45. 

46. 

47. 

48. 

49. 


1 + 1 


m — 1 m + 1 

1 _ 1 

Tfi — 1 m + 1 

a _ b 

CL — 6 CL b 
X . X 


+ 


X + y X - y 

4 . 3 


+ 


3 £ — 1 2 a; + 1 

cl + b 


66 . Find 


60. 

61. 

62. 

63. 

64. 

a — b 


6 

2 

5p -3 

4y - 1 

2 

3 

4x - 3o 

6 z + a 

4 

3 

3 x — 3y 

2 * + 2t, 

3b , 6 

5a + 10 

3a -3 

1 

5 

6 x — 2y 

30 a: — 9 y 


4 a 2 -9 b 2 (2 a + 3 b) 2 


Solution. —1. - a + fe . -g-~ 6 

4 a 2 — 9 b 2 (2 a + 3 6) 2 


a + 6 


a — b 


(2 a - 3 6)(2 a + 3 b) (2 a + 3 6)(2 a + 3 6) 
. (a + b){ 2 a + 3 5) - (a - 6)(2 a -3b) 

(2a - 3 6)(2a + 3 6)(2a +3 6) 








































FRACTIONS 


175 


4 _ (2 a 2 + 5 ab + 3 fr 2 ) - (2 a 2 - 5 ab + 3 b 2 ) 

(2 a - 3 6) (2 a + 3 6) (2 a + 3 6) 

(Complete this solution and check it.) 

Note 1. — Steps 1,2, and 3 of the rule on page 171 have been taken 
at once in Step 3 of this solution. The first fraction of Step 2 is equiva- 

lent t0 (a±b)^a±3b). the second tQ (o - ft) (2 a-3 ft) . Do 
L. C. D. L. C. 1_). 

this from now on. 

Note 2. — In Step 4, (2 a 2 + 5 ab 4-3 b 2 ) is (a + b)(2 a + 3 b), 
and (2 a 2 — 5 ab + 3 b 2 ) is (a — b) (2 a — 3 b). Always inclose such 
products in parentheses as in this Step 4. (Read the preceding sentence 
three times, — or more; enough to remember it.) 


a - 2 _ 

a -f* 2 

61. 

r + 4 r 4- 6 

a — 3 

a -J- 3 

r 2 — 3 r r 2 — r — 6 

x - 2 

a: + 2 

62. 

3 2 

(x + 5) 2 

1 

n. 

i 

to 

4 a 2 —9 6 a 2 —9a 

3 r + s 

9 r 2 + s 2 

63. 

£ ar 

i 

SO 

1 

v. 

CO 

9 r 2 — s 2 

a; 2 — 6 x 4- 8 x 2 — 16 

2m — 3 

, m — 9 

64. 

a 4 - 6 a — b 

4m - 2 

3 ra + 6 

2 a + 6 6 3 a + 3 6 

5 z — 4 i/ 3 X + 4?/ 

66. 

o + 6 a — 6 

5 x - y 

3 a: + y 

x 2 - 2 xy -\-y 2 x 2 — y 2 


66 . 


a + 5 


+ 


a -\- 3 


- a - 6 a 2 + 7a + 10 


fi7 a 4~ 3 b _ a — 3 6 

a 2 - 7 ab + 12 b 2 a 2 - ab - 12 b 2 

m + 2 _ m — 3 , m 4- 1 

ra 2 4- 2 m - 3 m 2 - 3 m + 2 m 2 + ^ - 6 


„„ c + 3d _ c + 2 d _ c - d 

69, 2c 2 + d- (? 3c 2 + 2cd-d 2 Gc 2 -5 cd + d 2 

_ a __ 1 _ b 

a 2 + 2 ab + b 2 a - b a 2 - 2 ab + b 2 


70 . 






































176 


ALGEBRA 


MIXED EXPRESSIONS 


122. An integral expression is an expression which does not 
have literal numbers in the denominators of fractions; as, 
a 2 + b 2 , or § a + b. 

An integral expression may be considered a fraction with 
denominator 1; thus, a + b is^the same as 9 + ^ • 


123. A mixed expression consists of one or more integral 

n 

terms plus or minus one or more fractional terms; as a 2 -f- -, 

a b a 

or 1 H-• 


124. Changing mixed expressions to fractions is like adding 

or subtracting fractions. 

Thus, 12f = 12 + - = — +3 = 

7 8 8 8 

Also, x + - ~ 3 = - + — = ? x + x -Z 3 = ~ 3 . 

’ 2 1 2 2 2 


EXERCISE 88 


Change the following mixed expressions to fractions : 


1. 10 

7. 2—!+! 

13. 

2 _ 2 x — 3 y 

3 

3 4 


2x + 3y 

2. 5 - - 

8. | + 2-i 

14. 

^ _ a — 5 x 

4 

5 6 


a — 2 x 

3 !+ 2 

9. 1 -f- —h 5 a 

15. 

5 a 2 - 19 z 2 

2 

2 


a 2 - 4 x 2 

4. |-3 

10. 2 5 ~ 2x 

16. 

2 + 2s3 

5 

3 


r 3 - s 3 

6. o + | 

11. o - 1 + A 

2 a 

17. 

5a: +2 

6 + a: 2 - 1 

6. c — | 

12. c - 4 + 

18. 

i -V 16 

3 

3 c 


a 2 — a 
















FRACTIONS 


177 


19. 2 + - 


a 




20. 3 -1 
b 



a 



25. —2-- - 1 
a — b 



29. a + & H--—r 

a — b 



s 


125. Multiplication and division of mixed expressions. 

In such examples, the mixed expressions are inclosed in paren¬ 
theses. Change the mixed expressions to fractions, doing what 
the signs direct; then multiply or divide as the signs indicate. 



Example. — Find 



Solution. — 1. 


/4(x + 1) - 1W16(* 2 - 1) +_7\ 

V * + i M * 2 -i ) 

4x + 4 — 1 ^ 16 x 2 - 16 + 7 
x + 1 x 2 - 1 

4 x + 3 x 2 - 1 
x + 1 * 16 x 2 - 9 
1 1 

(x - lH-s-Hr) 

-aH" -t * (4^H=^)(4 X - 3) 

1 1 


6 . 


x — 1 
4 x — 3 


Check. — Let x = 2. 


( 



1 1 

n^55 = n t 3_ = i 


3 ’ 3 3 ’ n 5 

1 5 



























178 


ALGEBRA 


EXERCISE 89 


( 4+ lX 8+ I) 

6. 


( 18 -l)( 2 4) 

7. 

(HX*+2S^u) 

G + ■)(!-') 

8. 

( 2+ IH 3 -g) 

c + .!jc-.:j 

9. 

( 6+ lM 2 ~f) 

( 2 + x-2y)( 1+ x -V y ) 

10. i 

(-SK‘t3 


“ ( 8 -rhM 3 -rf 3 ) 


13. (y-ii\+(iL _§A 

\x y ) \ 4 a; 2 t/ 2 / 

14. 

\Zb 2/ \6 6 4 a/ 

“■ 0 + j -’!+>/)• (■ 

”■ ( 8 +fe?M=>+^3) 

ie (l + li + \°\ . h - » + <«! 

\ x 2 - a? J \ 2 X -f 2 a) 

19 . (2 +—~)(2 - ^ W 2 + - -\ 

\ m — 3/ \ m + 2/ \ 4 m 2 — 16/ 





















FRACTIONS 


179 


20 


8 x 


('-5 

. /,_ !M _ Vi-_ 

\ x 2 + Qxy -f5 y 2 A x 


2 + 4 x + 3 
12 y 2 


){ 2+ ^H 3 + tH 


2 xy - 4 if 


2 + 5xy —14 y 2 / 


x+i/> 


126. A complex fraction is a fraction having one or more 
fractions in either or both of its terms; as. 


2 + 


and 


a — b a + b 

7 . ~ 

+ 


a — b a + b 


Since a fraction is merely an indicated division, complex 
fractions are merely another way of writing division of mixed 
expressions. a a 

Example. — Simplify a — a + P- • 

b _j_ a 

a — b a + b 

Solution. — This means: a. Find the difference in the numerator; 
b. find the sum in the denominator; c. find the quotient. 

The first two of these steps are carried along simultaneously below. 

a _ a a(a + b) — a{a — b) 

1 a — b a ~h b _ (a — b)(a + b) 

b , q b(a -j- b) H~ a( a b) 

a — b a + b (a — b) (a + b) 

1 1 

_ $ + ab — f + ab _ (q— 

2* — (p —(q—K&) ^6 + 6 2 4- a 2 — ^ 

1 1 

= 2ab - 

q 2 + 6 2 

(Check this solution by letting a = 2 and 6=1.) 

Note.— A complex fraction can be simplified by multiplying its 
numerator and denominator by the L. C. M. of the denominators of the 
fractions in its numerator and denominator. 






























co | oo 


ALGEBRA 


180 


Simplify: 

1 . 4 


5 - 


1 - 


2 . 


3. 


4 . 


o. 


8 . 


2 

3 

2 + - 
5 

*-£ 

3+ l 

5 -5 

- 1 


1 + 

3 

c 


1 


14 m 


i-i 

x 


EXERCISE 90 

-V 


y 


9. 


10 . 


11 . 


12 . 


13. 


14. 


y + 

X + 


2 i 

X L - 


2 x — - 
3 

2x4-- 

3 

111 

\ + y 


a , 5_6 
3 


x + 


y 


16. 


a - 

c 

a , 6 
6 a 


1 + 


17. 


18. 


19. 


i _«+«: 

b 2 

2 — — 


2 a 3 6 

3 6 2 a 

* - 13 + 19 

_X 

*- 3-19 


1 -i _ 21 

20_ y . _ £. 

2 _ 14 

y y 2 

a 


21 . 


a — b 


1 + 


a — b 

z 

22._ x y 


5-1 

x 2 — 

1 - * 

c 

4 

z + y 

7 m 2 a 

, 6 
a H— 

2 a i 


15 c 

oo o + 6 

oc 

6 

aO.-— 

a 


1 - 


24. 


a 4 - 6 
4 5 
r + 5 


_r_3jy 

3^ r 



































FRACTIONS 


181 


25. 


28. 


29. 


30. 


31. 


1 _ 1 

£_ y_ 

j_ 

xy 

x 


26. 


i+l 

x 


1 + ir— 


27. 


2d 2 
c -f- d 


c + 


d 2 


X + 3 


a; 2 + a; — 6 


z 2 + z + 1 - 


32. 


c — 2d 

1 

z — 1 


1 - 


x — 1 


y 

y 


m m + n 

y + 2 

y -3 

33. 

n m — n 

y 2 

y 2 

m , m + n 

y + 2 

y -3 


V 

n m — n 

3 

2 


2 + 1 - 1 

a — 2 

a — 3 

34. 

3 x 6 x 2 2 x 3 

1 - 

3 

2 _ 1 __3_ 

a — 2 


2 x 4 x 2 

2 a + 

1 - 2 a j 

3 a 

36. 

10(x 2 - y 2 ) 
a - 2 —9 y 2 

a - 

+ 

ICO 

i 

1 2(2 x + y) 

* + Sy 


SUPPLEMENTARY TOPIC 

127. For any fraction, three signs must be considered, — 
the signs of the numerator, of the denominator, and of the 
fraction itself. Certain rules for making necessary changes in 
these signs follow. 


Rule 1. — If the sign of one term of a fraction is changed, 
the sign of the fraction also must be changed. 


Thus: a. 


- 4 
+ 12 


1 . 

3’ 


but 


+ 4 
+ 12 



In this example, when — 4 is changed to + 4, the value of 
the fraction changes from — £ to + ^. 




































182 


ALGEBRA 



In this example, when + 12 is changed to — 12, the value 
of the fraction changes from — ^ to + i- 

In case it is desired to change the sign of a fraction, the sign 
of one of the terms of the fraction must also be changed. 

Rule 2. — If the signs of both terms of a fraction are changed, 
the sign of the fraction is not changed. 



Thus, 


In this example, — 4 is changed to + 4, and + 12 to — 12, 
but the value of the fraction remains — -J. 

The need of these rules is illustrated by the following example. 

Example. — Reduce to lowest terms--*• 


Solution. — 1. The terms of the denomin .tor, not being in the same 
order as those of the numerator, must be rearranged. 

fpk en x 2 — 9 _ _ x 2 — 9 _ _ 

12 + 2 x — 2 x 2 — 2 x 2 + 2 x + 12 * 

2. The negative coefficient of — 2 x 2 is inconvenient. Change the 
signs of the denominator by multiplying by — 1 . Then by Rule 1: 


x 2 - 9 


x 2 - 9 


- 2 x 2 + 2 x + 12 


2 x 2 - 2 x - 12 


3. 


( x —3)(x+3) 
2 (x* - x - 6 ) 


1 


4. 


— —ff)(x + 3) ? _ x -f 3 

2& -3)(x + 2) 2 (x + 2 ) 

1 


Check. — Let x = 1 . 

x 2 - 9 


1-9 _ - 8 _ 2 


12 + 2 x - 2 x 2 


12 + 2-2 12 3 




















FRACTIONS 


183 


128. There are two ways in which the signs of a numerator 
or of a denominator can be changed. 

a. If the expression is not factored, change the sign of every 
term by multiplying each of them by — 1. 

Thus, (- 1)(- 2x 2 -3x + 4) = 2x 2 + 3x-4. 

b. If the expression is factored, change the signs of all the terms 
in an odd number of factors of the expression. 

Thus: 2 X (- 3) X 4 = - 24, but 2 X ( + 3) X 4 = +24. 

2 X (- 3) X 4 = - 24, but (- 2) X (+ 3) X (- 4) = + 24. 

Similarly, consider ( x — y){y — z). 

y — z becomes —y + z or z — y when its signs are changed. 

Therefore {x - y){y - z) = - (x - y){z - y), for changing the 
signs of y — z changes the sign of the product. 

On the other hand, (x — y)(y — z) = + (y — x)(z — y) because 
the signs of x — y and of y — z have been changed. This means that 
— 1 has been used twice as a factor, and therefore the sign of the product 
is not changed. 

Do not confuse “ changing the signs of the terms of an ex¬ 
pression ” and “ rearranging the terms of an expression.” 

Thus, 2 x — 3 becomes 3 — 2 x when its signs are changed. 

2 x - 3 becomes - 3 + 2 x when its terms are rearranged. 

Note. — Changing the signs of the terms of an expression changes 
the value of the expression. Rearranging the terms of an expression 
does not change the value of the expression. 


EXERCISE 91 


Reduce to lowest terms : 


1 . 

2 . 

3 . 


x — 5 
15 - Sx 


4 b 2 — a 2 
y 2 — x 2 
4x - 4y 


6 + m — m 2 
m 2 + m — 12 
15 c — 10 d 
2 d? — cd — 3 c 2 
4 my — 3 mx 

9 x 2 -9 xy —4 y 2 














184 


ALGEBRA 


7. Reduce to lowest terms —- —~ 

2 xyz{z - y) 

• 

Solution. — 1. (y — z) differs from (z — y ) only in the signs of y 
and z. Change the signs of the numerator (§ 128, b). Therefore, 
change the sign of the fraction. (Rule 1.) 

1 

Then (s - y)(y - z) _ _ (x - y)&-—y) 

2 xyz{z - y) 2 xyz(z~ — y) 

1 

_ x — y 
2 xyz 


2. In case of a result such as that preceding, change the signs of 
x — y, and then change the sign of the fraction. 



II 

1 

3 

1 

3 

x — 

y ~ x . 


2 xyz{z — ?/) 

2 xyz ’ ~ 2 xyz 

8 . 

(a - 6)(6 - c) 

2(e - 6) 

12 . 

ab 2 + a 2 6 — ,2 a 3 
a 2 + a6 — 2 6 2 

9. 

1 

CO 

1 

3 

13. 

0 ~ 2/)0 2 + xy + ?/ 2 ) 

1 

to, 

1 

1 

co 

3 ?/ 2 - 3 r* 

10 . 

x 2 - y 2 

14. 

(a — 6)(6 — c)(c — a) 

(y - x) 2 

(b — a)(b — c)(a — c) 

11 . 

1 1 

3^ 

16. 

{x - yf 

{y - xf 

Find: 



16. 

Ho 

1 

l—^ 

CO 

l 

17. 

x 2 - 4x + 4 2 x + 5 

6 + a: — x 2 x — 2 

6x + 15 4 — x 2 


18 * 2 -l , 4 — x 2 3 x — 6 

' 4 - 2x x 2 - x - 2 2 - x - x 2 


^ 10 4- 5 m m 9—3 m 12 —3 m 2 
8m - 4 10 m 4- 5 8 m 2 —2 

on a — 2 a 2 1 + 4 a + 4 a 2 10 a — 10 a 2 
4 a 2 — 1 5 a 


2 + 2 a - 4 a 2 






























FRACTIONS 


185 


21. Find —£- 
a — b 


26 + a ^ 2 
b 2 — a 2 6 + a 


Solution. — 1. It is convenient to have the denominators and nu¬ 
merators arranged in descending powers of a. 

Therefore we write: 

3 , a + 2b _ 2 

o — 6 — o 2 + 6 2 o + 6 

This has the same value as the original expression, since the terms 
have been merely rearranged. No signs have been changed yet. 

2. In the second fraction, the denominator — o 2 + 6 2 is inconvenient 
since the coefficient of — o 2 is — 1. Therefore, change the signs of this 
denominator, and of the fraction. (Rule 1, page 181.) 

3 , 26 + o _2_ = 3 _ o + 26 _ 2 _ 

*' a - 6 6 2 - a 2 6+o a-6 a 2 - b* a + 6 

(Complete this solution.) 


2 x 

i 3 

24 2 - 

h x + 2 . 

a 2 - 25 

5 - x 

6 a — 9 

9 - 4 x 2 

1 

1 

26 4 

a - 13 

3z - 6 

1 

00 

1 

a 2 -5a 

10 a - 2 o 2 


_ 2 3 _ 4c_ 

3+c 3 — c c 2 — 9 

27 m + 1 - +_-- 

5 m 2 —7m 7 + 2 m —5m 


28. 


x + 2 . 2 - x . 16 

a:-2* 1_ 2+a: 4-z 2 


2 5 

^ x 2 — 5 x 6 15 - 2x - X 2 

on a + 3 , 3 a + 2 . 2 -_a _ 

3 a 2 + 7 a + 2 l+a-6a 2 2 - 3 a - 2 a 2 

Note. — Additional examples are to be found on page 432. 




































X. SIMPLE FRACTIONAL EQUATIONS 


129. Changing equations which involve fractions to equations 
which do not involve fractions. You have already done this 
in examples on page 14. You are now ready to do more difficult 
ones. 

Example. — Solve the equation 


3 x — 1 4 x — 5 


= 4 + 


7 x + 5 
10 


Solution. — 1. To, eliminate (get rid of) the fractions, multiply by 
the L. C. M. of the denominators, 20. 


2. Mso 

3. 

4. 

5. 

6 . 

7. D_ lfi 


20 . U _ 20 -( 4 * ~ 5 ) = 20 
i 5 


2 

4 + 20 


(7 rr + 5) 
X0 

5(3 * — 1) - 4(4 x - 5) = 80 + 2(7 x + 5). 

*. 15 x - 5 — 16 x + 20 = 80 + 14 re + 10. 

.*. 15 — x = 90 + 14 x. 

- 15 x= 75. 
x = - 5. 


Check. — Substitute — 5 for x in the original equation. 


Does 


16 


25 —At ~ 30 ^ 
5 10 ' 


Dofes —4+5 = 4 — 3? Yes. 


— 5 is the correct root 
of the equation. 


Note 1 . — The fractions can be eliminated by multiplying by any 
common multiple of 4, 5, and 10; for example, by 100. Naturally, we 
multiply by the least common multiple of 4, 5, 10. 

Note 2. — In checking from now on, try to do some of it mentally. 
Thus, in the check of this solution, — 16 is 3(— 5) — 1; — 25 is 
4(— 5) — 5. ' 

The process of changing the given equation to one which does 
not have any fractions is called clearing the equation of fractions. 

While the use of this expression is not necessary, its use is not ob¬ 
jectionable 'provided, you remember that u clearing the equation of frac¬ 
tions” is accomplished by multiplying both members of the equation 
by the least common multiple of the denominators. 

186 










SIMPLE FRACTIONAL EQUATIONS 


187 


EXERCISE 92 

Solve and check the following equations : 


1 x , 3 x _ 17 

9. r+16 - 

r + 4 

5 + 2 2 

6 

3 

2 5y _7y _1 

10. 3< + 5 4 

2 £ + 3_ ^ 

4 6 4 

8 

4 

o z + 5 , 0 2 + 16 

11 4x ~ 1 

H 

II 

1 

1 

3 ' 3 + - 4 

3 

2 12 

. w — 8,0 w + 12 

12. 6 f + 3 - 

o 

II 

LO 

1 

CO 

L 

6 7 

7 

3 

6 c - 6 . i _ 2 c + 4 

13 ^ Z ~ ^ : 

_2 + 1 11 

6 ' 8 +1 9 

3 

2 8 

m + 4 _ 13 m + 9 

14. 7 ^ + 6 

_ a 14 y + 1 

' 2 2 10 

3 

4 

a — 5 12 — a _1 

16. 122 + 2 

4z + 1 3 

* 4 6 2 

11 

2 

y> 

+ 

h-A 

II 

1 

1 

4^ 

16. * - 8 — 

5 + 5 _ 1 

8 2 

9 

2 2 

7x - 8 7 x + 6 

x — 5 4 x + 9 

17.- 1 — 

14 4x 

1 

II 


Solution. — 1. The L. C. M. of the denominators is 28 x. 


2. Ms 


3. 


U -4rz- 


14 x 

= 2§rxr 


-%■ 

2 x(7 x - 8) - 7(7 x + 6) = 14 x(x - 5) - 4(4 x + 9). 
(Complete this solution and check it.) 


(4 x + 9) 

^Xr 


Note 1 . — Observe Steps 1 and 2 of the solution. Use similar steps 
in the following examples. 

Note 2. — The result should be f. While you can check the solu¬ 
tion by substituting possibly your teacher will permit you to check 
such fractional results by going over the solution most carefully. 






































188 


ALGEBRA 


18 . 12 _ £ = i 


m 


19. 

20 . 

21 . 

22 . 


m 

3 6 

4 m 5 m 


5 


13 

20 


- 5 + 1=0 

2 a: a: 12 


8 = ± 
t 5 1 
3 _ _2_ 
y 5y 


_ 9 
5 

= 13 


28. 


23. 

24. 

25. 

26. 

27. 


2 r 


r 

5 5 + 6 


6 r + 1 _ 20 
3 r 3 
75 - 1 


28 
3 c 


5. 

14 - 2c 


__ 4 

5c 3 
6 a — 5 _ 9 a + 5 _ 25 
2 “ 3 18 a 

z—8 , 3 z — 6 


+ 


32 




X + 2 z 2 — 4 


Solution. — 1. The L. C. M. of the denominators is x 2 — 4. 
Multiply both members by x 2 — 4. 

.* + 2 o (*«2) . 1 o 

(x?-4)---= (xl —-4)--- 


2. (z!—4). 

3. 


(x- —-2) 

2(x + 2) - 5(« - 2) = 2. 
(Complete and check this solution.) 


Note 1. — As soon as possible, do Step 2 mentally. Always write 
Step 3, however, carefully inclosing the numerators in parentheses as 
is done in this solution. 

Note 2. — When solving fractional equations, the only root obtained 
may be a number which makes one or more denominators zero. Thus, 
in this example, if the root obtained is 2, the denominators of the first 
and third fractions become zero, and then these fractions do not have 
any meaning. (See § 68.) In such cases this root is only an apparent 
root and the “equation” is only apparently an equation, for a condi¬ 
tional equation is an equation only if there is a number which satisfies 
it. (See § 77.) 


29. 

15 = 5 

CO 

J-* 

1 

cn 

_ x 4- 2 


x — 3 

x + 2 

x — 5 

30. 

4 _ 3 

32 C + 7 

15 

m — 1 m — 2 

c - 5 

c - 5 























SIMPLE FRACTIONAL EQUATIONS 


189 


33. 


34. 


35. 


x — 5 _ x — 2 


36. 

CO 

1 

« 

1 

co| 

II 

rH 

x + 5 a: -f 2 


to 

1 

1 

O 

1 

to 

4 a: _ 3 a: 

1 

37. 

3 _ x -f 12 x 

to 

s 

1 

00 

1 

to 

K 

1 

00 

10 

a: -f 3 x — 3 x — 

12 X 2 + 3 X - 1 _ . 


38. 

5 x + 1 3 _ x 

3 x 2 + 2 x — 4 


2 x — 3 5 2 x — 3 

39. 91 - 

8/ 

+ 1 



40. 


41. 


42. 


43. 


o -1- 2 
2 r 

r — 4 
2 

< + 1 


+ 


a _ a + 20 
a 2 -4 
3 r 2 + 16 
r 2 - 16 
1 


a - 2 
r 

r + 4 
6 


2/ 2 + 5 y - 30 


2 * + 1 

+ 


f - 3 

y 


9 

46.f + |=-| 


/-2y-15 3 /+3 

44. 2^ + 3 . ? + J_ = * + i 

2 x + 1 3 x 
i + l 2 x 2 — 7 


y - 5 


9x — 1 
49. 


m 


47 


48. 


(Rini. — Review § 105, page 
147.) 

t 3 


3m + 4 7m— 4 

. 14 — £ 

50. 4x -- 14 


= 0 


3t + 4 4t + 1 
2 o + 3 2 a + 9 


61. 6 r -f 


a + 8 


63. 


3 a + 4 
5x 2 + 6 
x 2 -4 


62. 


a; + 1 
85 - 3r 
r 

c — 1 


64. 


c — 1 
3 _ 5a: + 3 

x — 2 a: +2 
y _ 5 - y = 15 


= 44 

_ 3 
2 


5 - 2 / 


y 














































190 


ALGEBRA 


EXERCISE 93 

1. One sixth of a certain number is 10 less than one half the 
same number. What is the number ? 

2. There is a certain number such that the sum of one half 
of it, one third of it, and one fourth of it is 3 more than the 
number itself. What is the number ? 

3. Separate 62 into two parts such that five ninths of the 
greater shall be 7 more than one half of the smaller. 

4. The sum of two numbers is 44. If the larger be divided 
by the smaller, the quotient equals §. What is the number ? 

6. One ninth of the result obtained by adding 12 to a certain 
number is one half the result obtained by subtracting 9 from 
the number. What is the number? 

6. One sixth of “11 more than a certain number/’ added to 
one third of “10 less than the same number,” gives 1. What 
is the number? 

7. What number added to both terms of the fraction f 
changes that fraction into one whose value is -|? 

8. What number added to the terms of the fractions f and 
f makes the resulting fractions equal ? 

9. A’s age is twice B’s age. If A’s age 4 years from now be 
divided by B’s age at that time, the quotient is f. What are 
their ages ? 

10. A and B are carrying loads of 40 pounds and 50 pounds 
respectively. How many pounds must be taken from A’s load 
and added to B’s so that A will have three fifths as much to 
carry as B ? 

11. The denominator of a certain fraction is 5 more than the 
numerator. If the denominator be increased by 1, and the 
numerator be decreased by 4, the value of the resulting fraction 
is -J. What is the given fraction ? 

12 . The numerator of a certain fraction is 9 less than twice 


SIMPLE FRACTIONAL EQUATIONS 191 

the denominator. If 1 be added to both terms of the fraction, 
the resulting fraction has the value -f. What is the fraction ? 

13. There are two consecutive even integers such that the 
larger is % of the result obtained by subtracting 3 from the 
smaller. What are they? 

14. A’s age is two thirds of B’s age. Ten years ago it was 
one half of B’s age then. What are their ages now ? 

15. Three fifths of a certain angle equals three fourths of the 
complement of the angle. (See Ex. 4, page 23.) What is 
the angle? 

16. There are three consecutive even integers such that one 
third of the first, plus one half of the second, plus one fourth of 
the third equals 15. What are they ? 

17. The perimeter of a certain triangle is 49 inches. The 
second side is J of “ 1 less than the first side ”; the third side 
is 4 of “ 1 more than the first side.” What are the sides of the 
triangle ? 

18. There are three consecutive integers such that one fifth 
of the largest is one fourth of the result obtained by subtracting 
1 from the smallest. What are they? 

19. Separate 58 into two parts such that the sum of one 
third of the greater and four sevenths of the smaller is 26. 

20. The denominator of a certain fraction exceeds its 
numerator by 22. If 7 be subtracted from both terms of the 
fractions, the resulting fraction has the value What is the 
fraction ? 

21 . One tenth of the supplement of a certain angle exceeds 
one half the complement of the angle by 1 degree. Find the 
angle. 

22. One fifth of the supplement of a certain angle exceeds 
four sevenths of the complement of the angle by 5 degrees. 
Find the angle. 


192 


ALGEBRA 


130. Work problems. If a man can do a piece of work in 
8 days, then, in one day, he can do one eighth of it, and in 3 days 
he can do three eighths of it. 

If a man can do the piece of work in x days, then, in one day, 

he can do - part of it, and in 5 days he can do 5 • - or - part of it. 
x XX 


EXERCISE 94 

1. If a man can plow a field in 12 days, what part can he 
plow in 1 day ? in 4 days ? in n days ? 

2. If a man can do a piece of work in y days, what part can 
he do in 1 day ? in 3 days ? 

3. A can do a piece of work in 6 days and B can do it in 
8 days. 

a. How much can A do in 1 day? in x days? 

h. How much can B do in 1 day ? in a: days ? 

c. How much can they do together in 1 day ? in x days ? 

4. A can do a piece of work in 8 hours and B can do it in 5 

hours. How long will it take them to do it if they work to¬ 

gether ? 

Solution. — 1. Let x = the no. of hr. it will take them to do it to¬ 
gether. 

2. Since A does 1 in 1 hr., he will do - in x hr. 

8 8 

Since B does l in 1 hr., he will do - in x hr. 
o 5 

/. together they will do § + - of it in x hr. 

8 5 

3. They complete the work in x hr.; that is, they do £$ or 1 times 
it in that time. 

••i+i- 1 - 

(Complete the solution.) 

Note 1. — Observe especially Step 3. 


SIMPLE FRACTIONAL EQUATIONS 193 


Note 2. — The essential facts of the solution can be charted. 


Person 

Time Alone 

Amount in 1 Hr. 

Amount in x Hr. 

A 

B 

8 hr. 

5 hr. 

l 

¥ 

1 

X 

■g" 

X 

Together 



or 1 



6. How long will it take A and B to do a piece of work 
together, if A can do it alone in 10 days and B in 20 days ? 

6. How long will it take A and B to do a piece of work 
together, if A alone can do it in 12 days and B in 9 days ? 

7 . How long will it take A, B, and C to do a piece of work 
together, if A can do it alone in 10 days, B in 12 days, and C in 
15 days ? 

8. A man can plow a certain field with his team in 8 days. 
His son can do it with their tractor in 2 days. How long will 
it take them to do it together? 

9. A can do a piece of work alone in 10 days and B in 15 
days. After A had worked x days and B (x + 3) days, they 
found that they had done £ of the work. How long had each 
worked ? 

10. In a factory, there is one machine which can turn out a 
certain number of articles in 8 hours, a second which can do it 
in 16 hours, and a third in 20 hours. How long will it take 
to fill a rush order for this number of articles, if all three ma¬ 
chines are worked at the same time ? 

11 . A certain machine requires 25 hours to complete a certain 
job. After it had been running 17 hours, it became necessary to 
complete the work on a machine which requires 40 hours for the 
same job. How long should it have taken to complete the work ? 


















194 


ALGEBRA 


131. Lever problems. A teeter board is one form of lever . 
The point which supports the lever is called the fulcrum; the 
parts of the lever to the right and left of the fulcrum are called 
the lever arms. 

If the weights L and R just balance, it is well known that, 
if R moves farther to the right, while L is stationary, then the 

right side goes down; and if R 
i L . ^ moves toward the fulcrum, then the 

right side goes up. Thus, the in¬ 
fluence of R upon the lever depends upon both the weight of R 
and its distance from the fulcrum. 

The influence of a weight upon a lever is called its leverage. 
It can be shown that the leverage of a weight is measured by the 
product of the weight and its distance from the fulcrum. 

Thus, if R weighs 50 pounds and is 4 feet from the fulcrum, its lever¬ 
age is 200; and, if L weighs 80 pounds and is 2 % feet from the fulcrum, 
its leverage is 80 X 2| or 200 also. The two will balance. 


This truth may be tested in the following manner. 

1. Remove a side and an end from a crayon box. Balance a stiff 
ruler on the edge of the box (the fulcrum). 

2. Place two pennies 6 inches from the fulcrum on the left side. 
Find where four pennies must be placed on the right side of the fulcrum 
to balance them. (It should be 3 inches to the right.) Notice that 
3 X 4 = 12 and that 6 X 2 = 12; that the 
leverages are equal. 

3. Find where 3 pennies must be placed, on 
the right, to balance the 2 pennies on the left. 

Find the leverage of the 3 pennies and com¬ 
pare it with the leverages in Step 2. 

4. Keeping the two pennies 6 inches to the 
left, place two others 2 inches to the right, and 2 four inches to the 
right. The ruler should be in perfect balance again. 

The leverage of the first two pennies on the right is 2 X 2 or 4; of 
the second two, is 2 X 4 or 8; 4 + 8 = 12. 



Example. — Suppose that the weights and distances in the 
figure on page 195 are: 









SIMPLE FRACTIONAL EQUATIONS 


195 


A. 40 pounds; distance 5 feet. B. 45 pounds; distance 4 
feet. C. 55 pounds ; distance 6 feet. D. 60 pounds; distance 
6 feet. 

Where must E be placed so that the lever will balance ? 

cab D E 


Solution. — 1. The leverages are: 

A. 5 X 40 = 200. C. 6 X 55 = 330. E. 50 X * = 50 x. 

B. 4 X 45 = 180. D. 6 X 60 = 360. 

2. .*. 200 + 180 + 330 = 360 + 50 x 

.*. 350 = 50 x, or 7 = x. 

That is, E must be 7 feet from the fulcrum. 

Rule. — To balance a simple lever: 

The sum of the leverages of all weights (forces) on one side 
of the fulcrum must equal the sum of the leverages of all weights 
(forces) on the other side of the fulcrum. 

EXERCISE 95 

1. A boy weighing 85 pounds sits 7 feet from the fulcrum 
and balances a boy who is sitting 6 feet from the fulcrum on the 
other side. What is the weight of the other boy ? 

2. C, weighing 105 pounds, sits 5 feet from the fulcrum on 
one side. Where must B, who weighs 75 pounds, sit, to balance 
him? 

3. A, B, and C weigh 6770, and 90 pounds respectively. 
If A sit£ 6 feet from the fulcrum, and B sits on the same side, 
4| feet from the fulcrum, where must C sit to balance them ? 

4 . A man is carrying two packages, by balancing them over 
his shoulder at the ends of a stiff rod, 4 feet long. The packages 
weigh 25 and 35 pounds respectively. How far from his shoulder 
should the 25-pound package be carried ? 

5. Two girls, weighing 75 pounds and 85 pounds respectively, 
wish to sit at the ends of a teeter board, 10 feet long. How far 
from the lighter girl should the fulcrum be placed ? 




ALGEBRA 


Ldditional rate, time, and distance problems. 


EXERCISE 96 

1. After reviewing § 85, page 111, give a simple arithmetical 
example involving rate, time, and distance. 

2. What are the rule and the formula for finding the: 

a. distance, when the rate and time are known? 

b. rate, when the distance and time are known ? 

<j. time, when the distance and rate are known? 

3. The rate of one train is f that of another. Let r repre- 
sent/tl eirate of the slower train. 

^present the rate of the faster train. 

^present the time each requires to go 125 miles. 

?peat Example 3 if the rate of the faster train exceeds 
that of the other train by 10 miles per hour. 

6. The rate of a passenger train is twice that of a freight 
train. It goes 200 miles in 2\ hours less time than the freight 
train requires for a trip of 150 miles. Find the rate of each. 



Solution. — While the “times” are unknown, the equation is clearly 
to be obtained from the relation between the time each requires for its 
trip. These “times” must be expressed. 


Let r = the rate of the freight train. 


Train 

Rate 

Distance 

Time 

Freight 

r m. per hr. 

150 m. 

159 hr. 
r 

Passenger 

2 r m. per hr. 

200 m. 

200 hr. 

2 r 


. 200 _ 150 
2 r r 


(Complete and check this solution.) 
















SIMPLE FRACTIONAL EQUATIONS 


197 


6. Work Example 5 again, letting t = the number of hours 
the passenger train requires for the trip. Represent the “ time ” 
of the freight train; obtain the expressions for their rates; form 
and solve the equation. 

7. The rate of a certain passenger train is twice that of a 
freight train. The passenger train travels 140 miles in 2 hours 
less time than the freight train requires for a trip of 105 miles. 
What are their rates ? 

8. The rate of a certain passenger train is 10 miles more per 
hoim^fehan the rate of an automobile. It takes the train -f- as 
nluch time for a trip of 140 miles as it does the automobile. 
What is the rate of each ? 

9. The rate of one automobile party exceeds that of another 
party by 7 miles per hour. It goes 176 miles while the slower 
party is going 120 miles. What are their rates ? 

10. The rate of a certain freight train is f that of a passenger 
train. It takes the freight train 1| hours longer than it does 
the passenger train to go 180 miles. What are the rates of the 
trains ? 

11. The rate of a certain passenger train exceeds that of a 
freight train by 10 miles per hour. Its time for a trip of 210 
miles is f that of the freight train for a trip of 100 miles. What 
are their rates ? 

River Problems 

On a river, the direction in which the water is flowing is called down¬ 
stream, and the opposite direction is called upstream. When going 
downstream, a boat is carried along by the current of the river and 
whatever force is exerted within the boat; when coming upstream, 
its progress is retarded by the current of the river. 

12. The rate of the current of a river is 3 miles per hour. 

a. What will be the rate downstream of some boys who are 
rowing at the rate of 5 miles per hour in still water ? b. Up¬ 
stream ? 


198 


ALGEBRA 


13. Repeat Example 12 if the boys are rowing at the rate of 
3 miles per hour in still water. 

14. How long will it take the boys in Example 12 to go: 

a. 16 miles downstream? b. 16 miles upstream? 

c. The round trip? 

16. a. Express the time for the boys of Example 12 to go d 
miles downstream, b. d miles upstream. 

c. Write the equation expressing the fact that the total time 
for the trip is 5 hours. Solve the equation. 

16. Some boys who can travel 10 miles per hour in their motor 
boat want to make a trip on a river whose current is 3 miles per 
hour. How far may they go if they have only 7 hours for the 
round trip ? 

17. The rate of the current of a certain river is known to be 
3 miles per hour. Some boys in a motor boat found that it took 
them as long to go 8 miles downstream as it did to go 3 miles 
upstream. What was their rate in still water ? 


SUPPLEMENTARY TOPICS 
133. Literal equations. 

Example. — Solve the equation ax 

Solution. — 1. 

2 . 

3. 

4. 


a 2 = — b(b + x) for x. 


5. Ha+6 


ax — a 2 = — b(b + x). 
ax — a 2 = — b 2 — bx. 
ax + bx = a 2 — b 2 . 
x(a + b) = (a + 6 ) (a - b). 

x= (z±Ml=3 i ora — 6 . 

(a + b) 


Check this solution by substituting (a — b) for x in the original 
equation. 

Note. — This solution means that x is equal to a — b regardless of 
the values of a and b. 



SIMPLE FRACTIONAL EQUATIONS 


199 


EXERCISE 97 

1. 7 x -f 10 a = 16 x — 17 a 6. a{a — x) = 6(6 -f- z) 

2. 3(x —2 b) = 8 x — 116 6. a(x — a) = 6(x — b) 

3. a(6x — a) = 6(a — bx) 7. c(x — c + 2 d) = d(x -f d) 

4. rs(l - 2 x) = 2 s 2 a: - r 2 8. r(x — s — r) = s(x — 2 s) 

9. rax -2 m 2 = mw — nx — n 2 

10. ra 2 (x — 1) = mx - 1 11. 2 ax + 6 6 2 = 3 bx + 4 a6 

12. 2 ax + d = 3 c — bx 

13. a 2 (x - 1) = 6(6 - 2 a + ax) 

14. a6(3 x + 4) = 2(a 2 + 3 6 2 x) 


16. b 2 = a(a + 6 2 x — a 2 x) 

16. c(x - 2) = d{x — 1) 

17. a(x — a) = 6(6 — a:) 


18. — = 1 + — 
x 2 

19 12 a _ 2x + 8a _ ^ 
x 5 x 

20 . “ =1 + 1 


22 . 


3x — 5a _ 9x — 7 a 
4 12 


3 x 


x _ 1 -f 2 ax 2 x -f 1 
2 ” 2 a a 2 


21 . 


m x m 
x - 3m 2 


2 x — 7 m 


24. 

26. 

26. 

27. 


= 1 


4 a 

x — 2 a 

6 5 

x—3c x — 5c 

a + 6 _ c 
2 x - 2 

x + a , 4 a 


28. 


29. 


2 x 


2 x 3 2 x 

3 x _ 

2 x + 3 c 2 x — 3 c 

6x + 5 a + _ 2 a 


x—3a x—3a 

2 x 2 - 15 c 2 
4 x 2 - 9 c 2 

3 x 


= 5 


x 2 


30 . 


2 x 2 — 2 ax ' x 2 — a 2 

9 2 = _ 

3 x — 5 a a—2a x — 3 a 


1 

























200 


ALGEBRA 


134. Deriving and using formulae. 


Example. — Derive a formula for the rate in terms of the 
amount , the principal, and the time in simple interest problems. 


Solution. — 1 . The formula for the amount is 


2. Mioo 

3. Sioo 

4. Dp/ 

5. 


A = P + 


Prt 


100 

100 A = 100 P + Prt. 


100 A - 100P = Prt. 

100 A — 100 P _ „ 

Pt 

or r = 100M - _P). 

Pt 


This formula enables us to find r when A, P, and t are known. Thus, 
if a man receives $3500 at the end of 6 years from an investment of 
$2400, what rate of simple interest has his money earned? 

Here, A = $3500, P = $2400, and t = 6. 


. _ 100(3500 - 2400) _ 1100 _ 7 6 . 

2400 X 6 144 ' ’ 

24 

that is, the rate is 7.6%. 


In the following list of examples, a number of formulae, taken 
from physics, chemistry, geometry, and engineering, are given. 
It is impossible to show in this text how these formulae are 
derived, as that calls for special knowledge of these various 
subjects. 

EXERCISE 98 


Solve each of them for the letters indicated: 

1. A = ab. a. Solve for a ; b. for b. 

2 . A = — • a. Solve for a ; b. for b. 

2 

3. C = 2 7rr. Solve for r. 

4. V = Iwh. a. Solve for w ; b. for h. 

6. V = -g- bh. a. Solve for b ; b. for h. 

6. F = f C + 32. Solve for C. 





SIMPLE FRACTIONAL EQUATIONS 


201 


7 . A - a ^ Jr -~ is a formula from geometry. 

2 

a. Find A when a = 15; b = 24; and c = 20. 

b. Find c when A = 550; b = 30; and a = 22. 

c. Solve the formula for a ; d. for b. 


8. ^4 


100 


o. Solve for P. 6. for £. 

9. T 7 = i + f is a formula from physics. 
a 

a. Solve it for L b. for a. 

10. mg — T = mf is a formula from physics. 
a. Solve it for T ; 6. for/; c. form. 


11 . w = ( —V is a formula from physics. 

\M -f m/ 

a. Solve it for m; b. for M. 

12. C = K a ] L i s a formula from physics. 

b — a 

a. Solve it for a ; b. for b. 

13. s = ° is a formula from physics. 

b — a 

a. Solve it for b ; 6. for a. 

14. h = Jc( 1 4- -zfe 0 is a formula from physics, 
a. Solve it for t ; 6. for k. 

15 i — L ~ is a formula from physics. 

Mt 

a. Solve it for L ; b. for M. 

16 w =- 5 a 0 — is a formula from engineering. 

i+— 

600 d 2 

a. Find w, correct to two decimals, when a = 18, l = 12* 
and d = 10. 

b. Solve the formula for o. 









(N CO ^ 


202 


ALGEBRA 


ad? • • • 

17. p = — + d is a formula from engineering. 


a. Find p when a = .56, d = \,t = §. 

b. Solve the formula for t. 


18. C =-is a formula from physics. 

R + - 

n 

a. Simplify it in the right member. 

b. Solve it for n ; c. for r. 

19. F = is a formula from physics. 

9 r 

а. Find F when m = 150, v = 25, g = 32, and r = 5. 

б. Solve it for r; c. for m. 

20. \ = - + i is a formula from physics. 

/ p q 

а. Find p when / = 30 and q = 40. 

б. Solve it for p ; c. for q. 


135. Equations involving decimals. 

Example. — Solve the equation .7 x — 1 = .36 x — .15. 
Solution a. — 1. .7 x — 1 = .36 x — .15. 

2. Mioo 70 x - 100 = 36 x — 15. 

This solution may be completed in the usual way, and checked by 
substitution. 


Solution b. — 1 . .7 x — 1 = .36 x — .15. 

.*. .7 x -.36 x = 1 -.15. 

.*. .34 z = .85. 

. D 34 x = 2.5. 

EXERCISE 99 


2.5 

?34)?85.00 
68 
17 0 
17 0 


Solve the following equations 

1. .5* + .5 = 6.1 - .3 a: 

2. .9 m- .75 = .3 m - 3.75 

3. .13 + .04 p = .11 p - .22 

4. .05 a - .12 = .16 - .09 a 


6. 2.1 -1.5 s = - .82 - .7 

6. 3 + 1.7 c = 2.7 c + 2.2 

7. .3(8 - 5 a) = .2 (a - 5) 

8. .3(a: - 2) = 5(.2 a: - 1.1) 





SIMPLE FRACTIONAL EQUATIONS 


203 


9. .5 = .3(x - 2) - (x - 6) 

10. 6(.2 - .3 x) + .3 = .3(4 x - 5) 

11 . .2 x - .03 * - .113 x = .01 + .161 

12 . .3 x - .02 - .003 x = .7 - .06 x - .006 

13 . .3(1.2 x - 5) = 14 + .05 x 

14 . ,7(x + .13) = .03(4 x - .1) + .5 

16 . 3.3 x - — - - — = .1 x + 9.9 
.5 

136. Meaning and use of ratio. The ratio of one number to 
another is the quotient of the first divided by the second. 

Thus, the ratio of 2 to 6 is f or £. It means that 2 is £ of 6. 

The ratio of a to b is the fraction ^ • 

b 

The first number (the antecedent) is made the numerator of 
the fraction, and the second (the consequent) the denominator. 

The ratio of two concrete quantities has meaning only if they 
are of the same kind. Their ratio is the quotient of their meas¬ 
ures in terms of the same unit of measure. 

Thus, the ratio of a segment 2 yd. long to a rectangle having 3 sq. 
yd. of surface is without meaning. 

The ratio of a fine 2 yd. long to one 10 ft. long is 6 ft. -v- 10 ft., or f. 

EXERCISE 100 

What is the ratio of: 


1. 2 to 12? 

6. 3 d. to 2 wk. ? 

2. 3 to 15? 

7. 6 in. to 3 ft. ? 

3. 1 pt. to 2 qt. ? 

8. 1500 lb. to 1 t. ? 

4. 3 oz. to 1 lb. ? 

9. 2\ to 5 ? 

6. 10 rd. to 1 mi. ? 

10. 3£ to4j? 

11. Out of an income 

of $2500, a man spends $400 for rent. 

What is the ratio of his rent to his income ? 



204 


ALGEBRA 


12. a. The altitude of rectangle R is 8 in. long and its base 
is 24 in. What is the ratio of its altitude to its base ? 

b. The altitude of rectangle S is* 9 in., and its base is 36 in. 
What is the ratio of its altitude to its base ? 

c. How do the ratios compare ? 

13. a. The altitude and base of one rectangle are 8 ft. and 12 
ft. respectively. What is their ratio ? 

b. The altitude and base of another rectangle are 11 ft. and 
33 ft. respectively. What is their ratio ? 

c. How do the ratios compare ? 

14. a. A purchased 12 t. of coal and B, 15 t. What was the 
ratio of the amounts purchased ? 

b. A paid $108 for his coal and B, $135. What was the ratio 
of their expenditures ? 

c. How do the ratios compare ? 

137. Meaning and use of proportion. If the ratio of two 
numbers equals the ratio of two other numbers, the four num¬ 
bers form a proportion. 

Thus, £ = Therefore 2, 4, 6, and 12 form a proportion. 

As a fractional equation, we read it two fourths equals six twelfths. 

As a proportion, we read it, 2 is to 4 as 6 is to 12; meaning 2 has the 
same relation to 4 as 6 has to 12. 

Similarly, the numbers a, 6, c, and d are proportional if - = -• 

b d 

Read it, a is to b as c is to d. 

a, b, c, and d are called the first, second, third, and fourth 
terms of the proportion. 

They are placed in the proportion in that order, a and d 
are the extremes of the proportion; b and c are the means of 
the proportion. 

A proportion is really only a very simple fractional equation 
and does not present any difficulties. 


SIMPLE FRACTIONAL EQUATIONS 


205 


m 


Example. — Find x from the proportion 

np 

Solution. — 1. 


2. M 


npx 


3. 

4. Dj» 


m 

np 


c 

nx 

V 


-nx 


jnpx-‘ — =jnpx- — 
jnp- ““ 

mx = cp. 

x=°2- 
m 


c 

nx 


EXERCISE 101 


Find x in each of the following proportions: 


1. 5 

_ 5 



3. 

7 

_ X 

x — 1 _ 5 

O. “ 

4 

3 




16 

~ 5 

2 3 

2.2 

_ 3 



4. 

9 

_ 3 

II 

CO 1 

1 

CD 

X 

5 




24 

X 

t + 5 6 

l. x - = 

b 

9. 

V = 

_ r 


u. ™=4 

13. —— -- 

a 

c 


q 

X 


x a 

a — x 

8. ™ = 

n 

10. 

r l 

_ r 


12. X - a -° 

14. * ■ 

X 

r 


SX 

~ t 


b 1 

a — x 


15. If it takes 75 gal. of road oil to cover 150 sq. yd., how 
many gallons will it take for 1525 sq. yd. ? 

Solution. — 1. In this example, it is implied that the amount of oil 
for 1525 sq. yd. is to the amount required for 150 sq. yd. as 1&25 sq. yd. 
is to 150 sq. yd. 

Note. — Observe, ain't of oil is to am't of oil as no. of sq. yd. is to 
no. of sq. yd. 

2. Let n = no. of gal. of oil required for 1525 sq. yd. 

. n _ 1525 

6 ' ” 75 150 * 

(Complete and check the solution.) 

16. If 20 acres of ground cost $3750, what should 35 acres 
cost at the same rate? 


gi s 







206 


ALGEBRA 


17. If 14 lb. of sugar can be bought for $1, what should 25 lb. 
cost at the same rate ? 


18. The weight of 18 cakes of ice was 5850 lb. What should 
be the weight of 30 cakes having the same average weight ? 

19. Two families together purchased 75 shrubs for $21.75. 
One family took 45 of them. How much should that family pay ? 

20. A train travels 130 miles in 4 hours. How many hours 
should it take a train to travel 250 miles at the same rate ? 

21. Separate a line 45 inches long into two parts which have 
the ratio 4 to 9. (Solve it in the two ways suggested below.) 


Solution a. Let x — shorter part; and .'. 45 — x = longer part. 
Solution b. Let 4 x = shorter part; and .". 9 x = longer part. 


22. Separate a line 84 inches long into three parts proportional 
a to 2, 3, and 4. (The method used in Solu¬ 
tion b above should be used in this case.) 

23. Separate 280 into four parts propor¬ 
tional to 2, 3, 4, and 5. 

24. In a triangle in which HE is parallel 
to BC, m: r = n: s. 



To test this truth: (a) measure m, n, r, and s ; 
(b) find the value of the ratio m : r and of n: s) ( c ) compare these two 
ratios. 


This truth may be tested in any triangle. It may be expressed 
thus: th<* upper segment on one side is to the lower segment on that side 
as the upper segment on the other is to the lower segment on the other. 


25. If AD = 7, DB = 4, AE = 8, find EC. 

26. If AB = 12, AD = 5, AC = 14, find AE. 
Hint. — Let AE = x, and CE = 14 — x. 

27. In the figure at the right, prove by 
measurement that 

AC _ BC 
AE DE 








SIMPLE FRACTIONAL EQUATIONS 


207 


28. Using the fact proved in Ex¬ 
ample 27, find the height of the tree, 
using the facts given in the figure at the 
right. 



29. Suppose that EF and AC are 
perpendicular to OC in the adjoining 
figure. Suppose that EF = 10 feet, 
OF = 12 feet, OC = 150 feet, and BC 
= 20 feet. Determine AB. 

30. Suppose that CD and AB are per- A( - 

pendicular to AE in the adjoining fig- ; 
ure; that AX = 5 feet, YB = 8 feet, ; 

AE = 750 feet, CE = 25 feet, and CD j 
= 30 feet. Find XY. c [--A 









XI. GRAPHICAL REPRESENTATION 


138. Graphical representation of statistics is illustrated in the 
graph below. 

Explanation of the Graph. — Opposite the word scale, observe 
the points marked Off, 10ff, 20ff, etc. This scale means that a segment 
(line, you may call it) from 0 to 10 represents 10 cents; one from 0 to 
30 represents 30 cents, etc. 

Opposite each month is a segment whose length, according to the 
scale, represents the price of eggs that month. Thus, the price in 
January was 40 cents. 


SCALE 

January 

.February 

March 

April 

May 

June 

July 

August 

September 

October 

November 

December 



Note 1 . — This graph is drawn on coordinate paper. You can 
make such graphs on ordinary paper by drawing parallel segments 
according to some scale. Thus, you can let £ in. represent Iff; then, 
in this example, the January price will be represented by a segment 
5 in. long, since 40 • in.) = 5 in. 


Note 2. — The segments can be drawn vertically or horizontally. 
Note 3. — Instead of drawing a segment, a narrow rectangle may 
be made. 

Thus, opposite January, 

The resulting graph is called a bar graph. 

208 













































































GRAPHICAL REPRESENTATION 


209 


EXERCISE 102 

1. In what months was the price lowest? During how many 
months did the price differ but little from this lowest price? 
What accounts for this low price during these months ? 


2. Determine from the graph the prices, and fill out the 
following table: 


Month 

Jan. 

Feb. 

Mar. 

Apr. 

May 

June 

July 

Aua. 

Sept. 

Oct. 

Nov. 

Dec. 

Price per doz. 














Represent graphically the statistics given in each of the 
following tables of statistics. 


3. The wholesale price of eggs on the first Tuesday of each 
month of a certain year in Chicago. 


Month 

Jan. 

Feb. 

Mar. 

Apr. 

May 

June 

July 

Aug. 

Sept. 

Oct. 

Nov. 

Dec. 

Price per doz. 

'"feL 

CO 

27 ?! 

19 ?! 

18 ?! 

18 ?! 

T—1 

16 ^ 

17 ?! 

22 ?! 

24 ?! 

27 ?! 

30 ?! 


4. The percentage of pupils studying algebra in one school, 
receiving each of certain grades. 


Grade 

95-100 

88-94 

81-87 

75-80 

In¬ 

complete 

Failed 

Per cent 

4 

18 

27 

26 

11 

13 

RECEIVING IT 






6. Tabular presentation of the average retail price of butter. 


During the 
Year 

1895 

1900 

1905 

1910 

1915 

1920 

Cost op 1 pound 

25 ?! 

CO 

<N 

29 ?! 

36?4 

36(4 

67 ?! 



























































































210 


ALGEBRA 


6. Tabular presentation of the population of the United 

States. 


Date 

1860 

1870 

1880 

1890 

1900 

1910 

1920 

Number op 

MILLIONS 

30 

39 

50 

63 

76 

92 

106 


7. Tabular presentation of the distance in which an auto 
can be stopped when going at certaip speeds. 


Speed in Miles per Hour 

10 m. 

15 m. 

20 m. 

25 m. 

30 m. 

Number op feet before 

it can be stopped 

9 ft. 

21 It. 

37 ft. 

58 ft. 

83 ft. 


8. The average temperature each month at Chicago. 


Month 

Jan. 

Feb. 

Mar. 

Apr. 

May 

June 

July 

Aug. 

Sept. 

Oct. 

Nov. 

Dec. 

Average 

temperature 

24 ° 

25 ° 

CO 

46 ° 

56 ° 

66 ° 

72 ° 

71 ° 

65 ° 

53 ° 

CO 

29 ° 


139. Variables. In each of the examples of the previous 
section there was a table of values which some changing number 
had under certain changing conditions. 

Thus, in Example 3, page 209, the “changing number” was the 
'price of eggs and the “changing conditions” was the month of the year. 

A changing number is called a variable. The variable is 
said to vary with or vary according to the changing conditions. 

The relation between the variable and the changing condition 
which caused the variation was expressed 

a. in tabular form, in the statement of the example; 

b. in graphical form, when you solved the example. 




















































GRAPHICAL REPRESENTATION 


211 


140 . Graphs involving positive and negative numbers. 

Example. — Draw the graph representing the following 
temperature readings. 


6 A.M. 

- 8° 

11 A.M. 

+ 7° 

4 P.M. 

+ 9° 

7 A.M. 

- 6° 

12 M. 

+ 8° 

5 P.M. 

+ 8° 

8 A.M. 

- 2° 

1 P.M. 

+ 9° 

6 P.M. 

+ 6° 

9 A.M. 

+ 2° 

2 P.M. 

+ 10° 

7 P.M. 

+ 4° 

10 A.M. 

+ 5° 

3 P.M. 

+ 10° 

8 P.M. 

0 



Explanation of this Graph. — On the horizontal line, called the 
horizontal axis, points representing 6 a.m., 7 a.m., etc., are placed at 
equal distances. 

On the heavy vertical line, called the vertical axis, + 5 means + 5°; 
— 10, means — 10°. Each space vertically represents 1°. This is 
called the unit. Segments drawn upward from the horizontal axis 
represent positive temperature; segments drawn downward from the 
horizontal axis represent negative temperature. 

Thus, the black vertical line above the point representing 11 a.m. 
is 7 units long. The temperature, then, was + 7°. 

The curved line, which is the graph proper, was obtained as follows. 
Instead of drawing lines at each hour like the one at 11 o’clock, the points 
were marked where these lines would end. (See the figure.) These 
points were connected by the smoothest possible curved line. 

This graph pictures the change in temperature from hour to hour 
during the day. 










































































































212 


ALGEBRA 


141. A graph may enable us to arrive at new information 
connected with the statistics being graphed. 



Example 1. — Referring to the figure above, what was the 
temperature at 6: 30 p.m. ? 

Solution. — 1. Point S on the horizontal line corresponds to 6 : 30 p.m. 

2. Go from S vertically to R on the graph. RS represents the tem¬ 
perature at 6 : 30 p.m. Go from R horizontally to + 5 on the vertical 
axis. 

3. the approximate temperature was + 5°. 

Example 2. — When was the temperature + 4° ? 

Solution. — 1. Without making any marks on the graph, start at the 
point on the vertical axis corresponding to + 4°. 

2. Go horizontally to the curved line, and from there go down to the 
horizontal axis. You arrive at a point two spaces to the right of 9; 
this point marks 9 : 40 a.m. 

3. .*. it w r as + 4° at 9 : 40 a.m. 

4. Also, a line from + 4° drawn horizontally crosses the graph at a 
second point which is above the point corresponding to 7 p.m. Hence, 
it was + 4° also at 7 p.m. 


EXERCISE 103 

1. Referring to the graph above, what was the approximate 
temperature at 6 : 30 a.m. ? at 8: 30 a.m. ? at 3 : 40 p.m. ? 













































































































GRAPHICAL REPRESENTATION 213 


2. Draw the graph representing the following table : 


6 A.M. 

+ 5° 

9 A.M. 

+ 15° 

12 M. 

+ 20° 

3 P.M. 

7 A.M. 

+ 8° 

10 A.M. 

+ 17° 

1 P.M. 

+ 18° 

4 P.M. 

8 A.M. 

+ 12° 

11 A.M. 

+ 19° 

2 P.M. 

+ 12° 

5 P.M. 


+ 4° 
- 2 ° 
- 5° 


3. If possible, get the temperature readings in your town 
for to-day, and represent them graphically. 


4. a. Complete the following cost table at 32*zf per pound 
of articles weighing 0 lb. to 5 lb. inclusive. 


Weight 

0 

1 lb. 

2 lb. 

3 lb. 

41b. 

51b. 

Cost 

o* 

32** 

64 

? 

? 

? 


b. Represent the above table graphically. 

Hint. — Let 1 space horizontally represent £ lb. Let 5 spaces ver¬ 
tically represent 20**. 

c. From your graph find the cost of 2\ lb.; 3f lb.; If lb.; 

4§ lb. 

6. Repeat Example 4 if the cost is 12^ per pound, for 0, 
2, 4, 6, 8, and 10 pounds. * 

6. a. The area of the rectangle with altitude 5 and base b 
is given by the formula A = 5 b. 

b. Make a table showing the area A, when b = 0, 2, 4, 6, 8,10. 

c. Represent this table graphically. 

7. The compound interest, to the nearest cent, on $1 in¬ 
vested at 4% compounded quarterly is given below. Express 
this information graphically. 


No. of Years 

1 

2 

3 

4 

5 

6 

7 

8 

9 

10 

Compound 

INTEREST 

.04 

.08 

1.13 

1.17 

1.22 

1.27 

1.32 

1.37 

1.43 

1.49 




































214 


ALGEBRA 


142. Definitions. In the figure below: 



a. lines XX' (read ex ex prime) and YY', drawn at right 
angles, are called the axes. 

b. XX' is called the horizontal axis or A'-axis. 

c. Y Y' is called the vertical axis or F-axis. 

d. Point 0, where the axes intersect, is called the origin. 

e. PR, perpendicular from P to XX', is the ordinate of P. 

f. PS, perpendicular from P to Y Y', is the abscissa of P 
(plural, abscissa r). 

g. The ordinate {PR) and abscissa (PS), together, are the co¬ 
ordinates of P. 

Since ORPS is a rectangle, PR = OS; therefore the ordinate 
of P equals OS. Similarly, the abscissa of P = OR. 

Distances up from XX' are positive; distances down from 
XX' are negative. Notice on OY the positive scale, + 1 
+ 2, +3, etc., and on OY' the negative scale, — 1, — 2, — 3, 
etc. 





































































































GRAPHICAL REPRESENTATION 


215 


Distances to the right from Y Y' are positive; distances to 
the left are negative. Again notice the positive and negative 
scales on XX'. 

Thus, the abscissa of P is -f 3; and the ordinate of P is + 4. P 
is called the point (+3, +4). This means that P is located 3 spaces 
to the right and 4 spaces up. 

When giving the coordinates of a point, the abscissa is always 
given first. 

Note. — The numbers of the scale should always be directly below 
the points of the horizontal axis, and exactly to the left of the points 
of the vertical exis. 

The scale need not be the same on the two axes. 

EXERCISE 104 

1. What is the abscissa of A? P? G? i/? 

2. What is the ordinate of E? B? C? D? 

3. What are the coordinates of A? E? B? G? H? 

4. On a similar sheet of paper, or on the blackboard, locate 
the following points: 

a. (3, 5) b. (- 4, 5) c. (-3,-6) d. ( + 4, - 3) 

e. (0, 7) /. (+5, 0) g. (0, - 10) h. (-8, 0) 

6. a. Locate points R: (2, 3) and S : (— 5, — 3). Draw RS. 

b. Locate points T: (— 3J, 1|) and W : (3, — 1). 

c. What are the coordinates of the point where RS intersects 

7TE? 

6. a. Locate C : (5, 4); D : (— 5, + 6); E : (— 7j, — §); 

F:( 2|,-2i). 

b. Draw CD, DE, EF, and CF. What figure is formed? 

c. Draw its diagonals. What are the coordinates of their 
point of intersection? 

7. a. Locate M: (+ 6, - 4) and N : (-6, + 2), and draw MN . 
b. What are the coordinates of the mid-point of MN ? 




216 


ALGEBRA 


143. Graphing the relation between two variables connected 
by a formula or equation. 

Example. — Represent graphically the relation between x 
and y, when x and y satisfy the equation 2 x + y = 4. 

Solution. — 1. Select values of x and determine the corresponding 
values of y. 

Thus, when x = — 5, 2(- 5) + y = 4; — 10 + y = 4; y = 14. 
when x = + 6, 2(6) + y = 4; 12 + y = 4; y = - 8. 

Similarly, 


When x = 

- 5 

- 3 

- 1 

0 

+ 2 

+ 4 

+ 6 

+ 7 

Then y = 

+ 14 

+ 10 

+ 6 

+ 4 

0 

- 4 

- 8 

- 10 


2. Use each pair of values so obtained as coordinates of a point. 
Represent each point by a small dot. 

Thus, place a dot at (- 5, + 14); at (- 3, + 10); etc. 

3. Draw the smooth line connecting these points. 



4. This is the required graph. Notice that it is a straight line. 

Note 1 . — The computation of Step 1 should be done mentally. 
Note 2. — This graph is usually called the graph of the equation 
2 x + y = 4. 







































































































GRAPHICAL REPRESENTATION 217 

EXERCISE 105 

1. a. What are the coordinates of C? Do they satisfy the 
equation ? 

b. Similarly, test the coordinates of B and D. 

2. a. Consider point (5, 5), not on the graph. Do these co¬ 
ordinates satisfy the equation ? 

b. Try the coordinates of any other point not on the graph. 

3. a. Express graphically the relation y = x — 3. 

Hint. — Let x— — 3, — 2, — 1, 0, 1, 2, 3, 4, 5. 

b. What does the graph appear to be ? 

c. Select a point not on this graph. Do its coordinates 
satisfy the given equation? 

4. a. Express graphically the relation x + 2 y = 6. 

Hi nt ' _ Follow the steps of the illustrative example. Let x = - 4, 

- 2, 0, 2, 4, 6, 8, 10. 

b. What does the graph appear to be? 

c. Select some point on the graph. Do its coordinates satisfy 

the given equation ? .... . » 

d. Select a point not on the graph. Do its coordinates satisfy 

the equation? 

5. Repeat Example 4 for the equation x — 2 y = 6. 

6. Draw the graph of the equation x + y = 5. 

7. Draw the graph of the equation x y — 8. 

144. Summary, a. The graph of an equation of the first 
degree having two variables is a straight line. 

For this reason these equations are called linear equations. 

6. The coordinates of every point on the line satisfy the 

\ The coordinates of every point not on the line do not 
satisfy the equation. 




218 


ALGEBRA 


145. Short method of drawing the graph of a linear equation 
having two variables. 

Since the graph is a straight line, and since a straight line can 
be drawn with a ruler as soon as two of its points are known, 
then 

Rule. — To draw the graph of a linear equation having two 
variables: 

1. Select a value for one variable. Determine the corre¬ 
sponding value of the other variable. These are the coordinates 
of one point on the graph. Locate it. 

2. Locate a second point in the same manner. 

3. Draw the straight line passing through these two points. 

4. Find a third point as in Step 1. If it falls on the graph, 
the work has been done correctly. 

Example. — Draw the graph of 4 x — 3 y = 6. 

Solution. — 1. Let x = 0. Then y = - 2. / Obtained mentally. 

2 * Let x = 3. Then y = 2. \ See Note, below. 

3. Locate the points and draw the graph. 



Check— Let x=Q. Then y=Q. Does this point lie on the line? Yes. 









































































































GRAPHICAL REPRESENTATION 


219 


Note. — The easiest value of z to use is x = 0, as in this example, 
for then — 3 y = 6 and y = — 2. 

In an equation like 3 x — 5 y = 11, when x = 0, — 5 y — 11, and 
y = — 2£. While this is correct, the point (0, — 2£) cannot be located 
easily. In such cases, try y = 0. Then 3 x = 11, and x = 3f. This 
is equally unsatisfactory. 

Try x= l, then 3 - 5 y = 11; - 5 j/ = 8; y = - If. Unsatis¬ 
factory. 

Try z = 2, then 6 - 5y = 11; - 5y = 5) y = - 1. This is a 
satisfactory point. 

Try z = - 3, then - 9 - 5 y = 11; -5y = 20; y = -4. 

Do not give up until you get at least two satisfactory pairs of numbers 
which satisfy the equation. If necessary, use fractional values of the 

variables. 


EXERCISE 106 

of each of the following equations, each on a 
of coordinate paper. 


Draw the graph 
separate piece 

1. x - y = 5 

2. 2 x - y = 3 

3. 3z = 4 

4. \ x - y = 5 
6. 2 y = 6 - x 


6. y — 3 x = 0 

7. 4 x + y = 0 

8. 4 x + 5 y = 20 

9. 3 x - 2 = y 

10. 3 x + 5 y — 15 = 0 


Supplementary Examples 

11. y = x + 2.5 14. V = 5.3 - .7 x 

12. y = 3 x - .75 15. $ * - $ J/ = 5 

13. 2/ = .5 x + .8 (.Hint. — M 6 .) 

16. The sum of two numbers is 9. Letting x and y represent 
these numbers, form the equation expressing the fact that their 
sum is 9. Then draw the graph of the equation. 

17. One number exceeds twice another number by 5. Ex¬ 
press these numbers by x and y ; form the equation; draw its 



220 


ALGEBRA 


146. Two linear equations having two variables. 

A solution of an equation having two (or more) variables 
consists of a value of each which, together, satisfy the equation. 

Thus, x = 1, y = 4 is a solution of x + y = 5, since 1+4 = 5. 

Also, x— — 8, y = 13 is a solution since — 8 + 13 = 5. 

Note. — Be careful not to confuse the word solution as just used 
with the word “solution” referring to the process of solving a problem 
or an example. 

Example. — Draw upon the same set- of axes the graphs of 
f 2 x — y = 4 
\2x + 3 y = 12 


Solution. — 1. Draw the graphs of the two equations. 


(1) 2 x — y = 4 

(2) 2 z + 3 ?/ = 12 

Point A 

When x = 0, y = — 4 

Point D 

When x — 0, y = 4 

Point B 

When x = 2, y = 0 

Point E 

When x— 6, y = 0 . 

Point C 

When x = 4, y = 4 

Point F 

When x = — 3, y = 6 

















































































































GRAPHICAL REPRESENTATION 


221 


2. Recall that the coordinates of every point on a graph satisfy the 
equation of that graph. (See § 144 b, page 217.) 

Observe point G. Its coordinates are x — 3, y — 2. 

(3, 2) satisfy 2 x - y =' 4, since 2 • 3 *- 2 does equal 4. 

(3, 2) satisfy 2 x + 3 y = 12, since 2-34-3*2 does equal 12. 

/. the coordinates of G satisfy both equations. 

3. Since no other point lies on both lines, no other pair of numbers 
can satisfy both equations. 

(3 ? 2) is the only pair of numbers which will satisfy both equations. 

4. These equations are called independent equations because each 
has solutions which are not solutions of the other; they are also simul¬ 
taneous equations (a special kind of independent equations) because 
they do have one common solution. 

Note. — On this graph, observe the points marked A, B, C, D , E, 
and F. Also observe that one line is marked (1), and the other (2), 
meaning that these lines are the graphs of equations (1) and (2) re¬ 
spectively. 

Rule. _ To determine graphically the common solution of 
two linear equations having two variables: 

1. Draw the graphs of the two equations upon one set of axes. 
(Use the form of page 220.) 

2. Determine the coordinates of the point which is on both 
graphs. This is the common solution. 

3. Check the common solution by substituting it in both 
equations. 


EXERCISE 107 


Determine graphically the common solution of: 


f x + y = 6 
1 x - y = 2 
f 2x + y = 9 
1 3z - y = 1 
f*+2y-ll 
\ Sx - y = 5 


j 3 x + y - 3 
| 2 £ - 3 y = 13 
( x + y = - 2 
[3x + 5y = - 2 
\ 6 x — 5 y = —1 
\.2 x + 3 y = 9 




222 


ALGEBRA 


7. a. Draw the graphs of 2 x — y = 4 and 6 x — 3 y = 12. 
6. What happens ? How many common points do the graphs 

have? How many common solutions , then, do the equations 
have? 

c. Such equations are called dependent equations. 

8. In the same way, study equations x + 2 y = 6 and 
2 x + 4 y = 12. 

9. a. Draw the graphs of 3 x — y = 5 and 6 z — 2 y = 18. 
6. What kind of lines are these graphs ? Do these equations 

have a common solution? 

c. Are these equations independent? (See Solution, Step 4, 
page 221.) 

d. Are these equations simultaneous ? 

e. These equations are inconsistent equations, another special 
kind of independent equations. 

10. Study in the same manner the equations 2 x — 3 y = 6 
and 4 x — 6 y — 30. 

Determine graphically what kind of equations each of the 
following pairs is. If they are simultaneous, determine their 
common solution. 


u f7*-2y-31 

16 . |4* + 3y - -1 

1 4 a; - 3 t/ = 27 

1 8 z + 6 ?/ = —2 

12. 3^ + 72/ =4 

16. (4*-32/ =5 

1 3 x + 7 j/ = - 15 

l8z-6y = 17 

i3. | 3x + y = 9 

17 f x — 6 y = 2 

l 5 s + 4 y =22 

1 — 8z + 3?/ = 29 

14 \3x + y = n 

18 ]ix — 2y = \ 

\ 5 x - y = 13 

1 6 z +4 7/ = 19 


Note. — Sometimes it is difficult to find the common solution 
accurately by this graphical method. Example 18 may have caused 
you some difficulty. In the next chapter, other methods of solution 
are taught. 


GRAPHICAL REPRESENTATION 


223 


147. Summary. 

a. Dependent equations are two equations having the same 
two (or more) variables, such that every solution of each equa¬ 
tion is a solution of the other. They have the same graph. 

(See Examples 7 and 8, page 222.) One equation can be changed 
into the other by multiplying both sides of the equation by the same 
number. 

b. Independent equations are two equations having the same 
two (or more) variables, such that each equation has solutions 
which are not solutions of the other. 

There are two kinds of independent equations. 

1. Simultaneous equations are two independent equations 
which have one (or more) common solutions. 

If they are linear equations, their graphs are two different 
straight lines which intersect at one and only one point. (See 
Example, page 220.) They have only one common solution. 

2. Inconsistent equations are two independent equations 
which do not have any common solution. 

If they are linear equations, their graphs are two parallel 
straight lines. (See Examples 9 and 10, page 222.) 

EXERCISE 108 


Supplementary Examples 

1. Draw a bar graph (page 208) representing the strengths 
of various woods. 


Wood 

Poplar 

Yellow 

Pine 

White 

Pine 

Red 

Fir 

Walnut 

White 

Oak 

Maple 

No. OF LB. TO 

EACH 8Q. IN. 

7,000 

11,000 

15,000 

10,000 

12,000 

14,500 

10,500 


Suggestion. — Let \ in. = 1000 lb., or let 4 spaces = 1000 lb. 















224 


ALGEBRA 


2. Make a table like the one in Example 7, page 213, giving 
the simple interest on $1 at 4% in 1, 2, 3 . . . 10 years. Draw 
the graph for the table you have made. 

3. Table of distances through which an object, falling freely, 
falls in various times. 


In 

0 sec. 

1 sec. 

2 sec.- 

3 sec. 

4 sec. 

5 sec. 

It falls 

Oft. 

16 ft. 

64 ft. 

144 ft. 

256 ft. 

400 ft. 


Represent this relation graphically. 

Hint. — Let 1 space vertically represent 8 feet; 5 spaces horizontally 
represent 1 second. 

4. Table of relation between Centigrade and Fahrenheit 
readings. 


Centigrade 

— 15° 

- 5° 

0° 

10° 

20° 

0 

O 

CO 

0 

C 

Fahrenheit 

5° 

23° 

32° 

50° 

68° 

86° 

104° 


a. Represent this relation graphically. 

Hint. — Horizontally, let 1 space represent 1° Centigrade. Ver¬ 
tically, let 1 space represent 2° or 4° Fahrenheit. 

b. From your graph, find the Fahrenheit temperature corre¬ 
sponding to 5° Centigrade; to — 10° Centigrade; to 80° Fahren¬ 
heit. 

5. Draw the graph of 3 V — 2 W = 6. 

Hint. — Let the horizontal axis be the F-axis; and the vertical axis 
be the TF-axis. 

6. S = 1.5 C + 2 is the formula for the selling price of an 
article costing C dollars, so that the gain is $2 more than 50% of 
the cost. 

a. Draw the graph of this equation, letting the horizontal axis be the 
C-axis. 

b. From your graph, find S when C = 5.5. 




























XII. SIMULTANEOUS LINEAR EQUATIONS* 


148. A linear equation or first degree equation having two 
or more unknowns is one in which no unknown appears in the 
denominator of a fraction, and in which the sum of the ex¬ 
ponents of the unknowns in each term is only 1. 

Thus, x + y = 10 is a linear or first degree equation. 

But xy + x = 10 is not a first degree equation since the sum of the 
exponents of x and y in the first term is 1 + 1 or 2. 

Also ^ -f y — 10 is not a first degree equation in x and y because x 

appears in a denominator. When the equation is cleared of fractions, 
1 + xy = 10 x, and this again is of degree two. 

149. A solution of an equation having two or more unknowns 
is a set of values of the unknowns which together satisfy the 
equation. 


Thus, in the equation x + y = 10, if x = 3, then 3 + y = 10, and 
therefore y = 7. x = 3, y = 7 is a solution of x + y = 10. 

Similarly, in the equation x + y = 10 


If X = 

1 

2 

4 

7 

10 

15 

- 2 

etc. 

then y = 

9 

8 

6 

3 

0 

- 5 

+ 12 

etc. 


Each of these pairs of numbers is a solution of x + y = 10. 


An equation with two, or more, unknowns has an infinite 
number of solutions. For each value of one unknown, there 
is a corresponding value of the other unknown. As one un¬ 
known varies (changes in value), the other also varies. The 
unknowns, therefore, are called variables (changing numbers). 

Therefore these equations are called, preferably, equations 
having two, or more, variables. 

* If you studied Chapter XI, you have had the definitions that ap¬ 
pear in §§ 148-151. 


225 
















226 


ALGEBRA 


150. Two linear equations having two variables are inde¬ 
pendent if each has solutions which are not solutions of the 
other. 

Thus, x + y = 10 has the solution x = 2, y = 8. 

2 x + y = 14 does not have the solution x = 2, y = 8, 
since 2 • 2 + 8 = 12. 

Therefore, x + y = 10 and 2 x + y = 14 are independent. 

They are said to form a system of independent equations. 

Two independent equations often do have common solutions. 
They are then called simultaneous equations. 

Thus, x + y = 10 has the solution x = 4, y = 6. 

Also, 2 x + y = 14 has the solution x = 4, y = 6. 

Simultaneous linear equations have only one common solu¬ 
tion. 

151. To solve a system of simultaneous equations is to find 
their common solution. 

One way to find the common solution of two equations would 
be to write down different solutions of each, hoping to obtain 
the common solution by trial. 


Thus, x + y = 10 x 

1 

2 

3 

4 

5 

6 

7 

8 

etc. 

has the solutions y _ 

9 

8 

7 

6 

5 

4 

3 

2 

etc. 


x - y = 2 

x = 

l 

2 

3 

4 

5 

6 

has the solutions 

y = 

- l 

0 

1 

2 

3 

4 


x = 6, y = 4 is the common solution. 


Much better ways to find the common solution are given 
below. 

152. Elimination by addition or subtraction. 

Example 1. — Solve the system / ^ X , ^ ^ 

\ 7 a: -f- 41/ = 2 


a) 

( 2 ) 




























SIMULTANEOUS LINEAR EQUATIONS 227 


Solution. — 1. M 4 * (1) 

20 x - 12 y= 76. 

(3) 

2 . M 3 ( 2 ) 

21 x + 12 y = 6 . 

(4) 

3. Add (3) and (4) 

41 x = 82. 

(5) 

4. D 41 (5) 

x = 2. 

(6) 

5. Substitute this value of x in (1) 10 — 3 y — 19. 

(7) 


— 3 y = 9, or y = — 3. 

(8) 

6. The common solution is 

x = 2, y = - 3. 


Check. — Substitute in (1) 

10 + 9 = 19. 


Substitute in (2) 

14 - 12 = 2. 



Notice. — 1. That the coefficients of x and y were not the same in 
equations (1) and (2). 

2. That equations (1) and (2) were multiplied by 4 and 3 respectively 
to make the coefficients of y the same except for sign, in equations (3) 
and (4). 

3. That the terms 12 y were made to disappear in Step 3 by adding 
equations (3) and (4). This is allowable because y represents in (3) 
and (4) the common value of y which we are seeking, y is said to be 
eliminated. This is an example of elimination by addition. 

4. If, in equations (3) and (4) the signs of 12 y were the same (for 
example, both -f) then, in Step 3, we should have subtracted, in order 
to eliminate 12 y. 

Rule. — To solve a system of two simultaneous linear equa¬ 
tions having two variables by the addition or subtraction method 
of elimination: 

1. Multiply, if necessary, both the first and second equations 
by such numbers as will make the coefficients of one of the 
variables of equal absolute value. 

2. If the coefficients have the came sign, subtract one equation 
from the other; if they have opposite signs, add the equations. 

3. Solve the equation resulting from Step 2 for the other 
variable. 

4. Substitute the value of the variable found in Step 3 in any 
equation containing both variables, and solve for the remaining 
variable. 

6. Check the solution by substituting it in both of the original 
equations. 

* Read this “multiply both members of equation (1) by 4.” 


ALGEBRA 


i 


228 


Historical Note. — Little progress was made in solving linear equa¬ 
tions having more than one unknown until the latter part of the 15th 
century, although mathematicians before that time had considered 
such problems. After Stifel and Stevin had introduced somewhat 
simple notations for several unknowns and their powers, definite methods 
for solving equations of the first degree with two unknowns were de¬ 
veloped. Johannes Buteo, a French monk (1492-1572), solved equa¬ 
tions with three unknowns in a text on algebra which appeared in 1559. 


EXERCISE 109 

Solve the following systems of equations by the addition or 
subtraction method of elimination: 



SIMULTANEOUS LINEAR EQUATIONS 


229 


153. Elimination by substitution is a second common method 
of solving simultaneous equations. 


Example. — Solve the system of equations 
f 7 x - 9 y = 15 
1 8 y - 5 x = - 17 

Solution. — 1. Solve equation (1) for x in terms of y: 


7 x = 15 + 9 y ; :. x = 15 + 9 J/. 

2. Substitute this value of rc in equation (2). 



Then 8 y - 

5(15 + 9 y) 
7 

= 

- 17. 

3. 

M 7 (4) 

56 y - 

• 5(15 +9i/) 

= 

- 119. 

4. 


.*. 56 y 

- 75 - 45 y 

= 

- 119. 

5. 

C. T.; 

At s 

ll y 

= 

- 44. 

6. 

Dn 


y 

= 

- 4. 

7. 

Substituting this value 

i of y in equation 

i(D, 



7 x + 36 = 

15; 7 x = - 

21 

; x = 

8. 


the common 

solution is x 

= 

~ 3, y 


( 1 ) 

( 2 ) 

(3) 

(4) 


Check. — In (1), does - 21 + 36 = 15? Yes. 

In (2), does - 32 + 15 = - 17? Yes. 

Note. — In Step 3, when multiplying + 9 y) ^ ^ we 0 ^ ta j n 
7 • 6(15 + 9 j/), or 5(15 + 9 y y 


Rule. — To solve a system of two simultaneous linear equa¬ 
tions having two variables by the substitution method of elimi¬ 
nation : 

1. Solve one equation for one variable in terms of the other 
variable. 

2. Substitute for this variable in the other equation the value 
found for it in Step 1. 

3. Solve the equation resulting in Step 2 for the second 
variable. 


(Continued on page 230.) 






230 


ALGEBRA 


4. Substitute the value of the second variable, obtained in 
Step 3, in any equation containing both variables and solve for 
the first variable. 

6. Check the solution by substituting it in the original equa¬ 
tions. 

Historical Note. — The earliest use of the Substitution Method of 
Elimination in print, of which we have any record, is in Newton’s 
Arithmetica Universalis, in 1707. 


EXERCISE 110 



! 13 z - 12 t = 35 
“• 1 x = 9 f 



SIMULTANEOUS LINEAR EQUATIONS 


231 


164. Simultaneous fractional equations. 


Example. — Solve the system of equations 


x + 3 y+4 

x(y — 2) - 2/(x - 5) = - 13 


( 1 ) 

( 2 ) 


Solution. — 1. Simplify equation (1) 


x + 3 y + 4 

M(ap + 3)(if+4> 7(?/ +4) — 3{x + 3) = 0. 

7y + 2S-3x-9 = 0. 

7 y - 3 x + 19 = 0, or | 3 x - 7 y = 19 | (3) 

2. Simplify equation (2) 

x(y - 2) - y(.x - 5) = - 13. 
xy — 2x — xy-\-5y = — 13. 

- 2 x + 5 y = - 13, or | 2 x - 5 y =~U ] (4) 

3. Therefore equations (1) and (2) become equations (3) and (4). 

3 x - 7 y = 19. (3) 

2 x — 5 y = 13. (4) 

4. Ms (3) 15 x- 35 y = 95. (5) 

M 7 (4) Ux-35y= 91. (6) 

5. (5)-(6) x= 4. 


6. Substitute in (4) 8 — 5 y = 13; — 5 y = 5; y = — 1. 
Check by substituting in (1) and (2). 


Equations (3) and (4) of this solution are called the standard 
form of equations (1) and (2). 

Rule. — To solve complicated simultaneous linear equations 
having two variables: 

1. Reduce the equations to standard form by clearing of 
fractions (when necessary) and simplifying. 

2. Solve the resulting system of equations by either method 
of elimination. 

3. Check by substitution in the original equations. 











232 


ALGEBRA 


EXERCISE 111 


Solve the following systems, 
traction method of elimination, 
method: 

'3_x _ 4j/ = _ i 

2 3 

2 x _ y _ 7_ 

3 4 12 

a +3 _6 = _ 3 

2 

3 

’ 2 


1. 


2. 


4 2 

6 +3 . 


2 

3d - 

4 c + 
1 


4 

c - 4 
6 

2d - < 


13 


: = - 7 


m — 1 

w + 6 

m — 5 _ 

2 

71 + 1 

= 3 

r .9 _ 

5 

6 4 

12 

3 + 2r 

7+35 

5 

10 


_ 1 

2 

2 

4 a; + 1 

1 

II 

6 

4 

[32-1 

2w + 1 

4 

2 

3 z + 1 

2 w - 1 

2 

3 


= 1 


some by the addition or sub- 
and some by the substitution 


2 2+5_1 

2 — 5 2 

4 — 2 — s _ 2 
5 


10 . 


11 . 


12 . 


13. 


14. 


2 2 

A + B -2 _1 
A — B 5 

i+25-3 _1_ 

i - 35 8 

3 a- - 4 y _ x + 8 y _ q 
2 5 

2 a: + y 4a; - 3 y _ q 


5 9 

w- 2 2 + 2 


W + 1 2 + 1 

w — 3 2 + 5 


w — 4 2 + 4 

a -36 — 1 _ j 
2a - 36 - 1 
5 - a + 66 = x 


7 a — 9 6 — 1 
X-Y X + Y 


4 

5 

2 X - Y _ 

2x+y 

6 

3 

32-1 

2 r — 1 

2 

i 

3 2 + 1 

2 (r + 1) 

2 

3 


= 3 


= 0 






















































SIMULTANEOUS LINEAR EQUATIONS 


233 


w 

c 

1 


3 a -f- 5 b 

4 a _ 17 


15 

10 

6 

1 Q 

2 

3 6 


3 4- 2 w 

_ 1 

lo. 

a — 3b . 

2a + b _ 

1 

7 - 

-3c 

2 


14 

7 

10 


1 

2 


f 2 p + q 

1 

CO 

<5 

II 

. 7 

4 y 

— X 

4 y + x 

19 { 

2 

5 

' 15 


1 

3 


P+ 3 ? _ 

CO 

1 

II 

2 

8 y 

- 5 

8 x — 5 


5 

7 

15 

M 

- N - 

-5 3 M-N 


5a; + 4 y 

CO 

H 

1 

to 

II 

. 11 


4 

16 

20. < 

2 

3 

' 10 

M 

+ 5 

M +N _ 11 

* + 4 y | 

1 

II 

9 


155. Dependent equations. 

Review the definition given in § 150. 

Dependent equations are two equations which have exactly 
the same solutions. For all practical purposes they are the 
same equation. One can be obtained from the other by mul¬ 
tiplying it or by dividing it by the same number. 

Thus, x + y = 10 and 2 x + 2 y = 20 are identical equations. 
x = 3^ = 7 is a solution of each. Any other solution of the one is 
also a solution of the other. 

If a system consists of two dependent equations, the system 
cannot be solved, because actually there is only one equation. 

^ , I x + y= 10 (1) 

Thus, consider j 2 x + 2 y = 20 (2) 

If we try to eliminate by substituting (10 — x) for y in (2), we get 
2 x + 20 - 2 * = 20, or 0 = 0. 

There is no opportunity to determine x. 

If you studied § 147, review it at this time. 

156. Inconsistent equations are two independent equations 
which do not have any common solution. 

Thus, x + 2 y = 7 and 4 x + 8 y = 40 are inconsistent. 

If 4 x + 8 y = 40 is divided by 4, we get x + 2 y = 10. 

Clearly, if x -h 2 2/ = 7, x+2y cannot = 10. 




























234 


ALGEBRA 


Notice that the coefficients 4 and 8 of the second equation are 4 
times the corresponding coefficients of the first equation, but that 40 
is not 4 times 7. This is characteristic of inconsistent equations. 

If we try to solve the system 

f x + 2y = 7 
\ 4 z + 8 2/ = 40 

by the subtraction method, we get 
( 4: x + 8 y = 28 
\ 4 x + 8 y = 40 

Subtracting, 0 = — 12 

This is impossible. 

157. When solving problems by means of two or more literal 
numbers, as many independent equations must be obtained 
from the conditions of the problem as there are letters used. 

Note. — Most problems that can be solved by means of equations 
having two unknowns, can also be solved by an equation having only 
one unknown. Nevertheless, use two unknowns when solving the 
following problems. 

EXERCISE 112 

1. Find two numbers whose sum is 20 and whose difference 
is 7. 

Solution. — 1. Let l = the larger number and s = the smaller 
number. 

(Complete the solution.) 

2. Separate 39 into two parts such that 4 times the smaller 
exceeds the larger by 6. 

Solution. — 1. Let l = the larger part and s = the smaller part 
of 39. 

2. I + s = ? 

3. Also 4 s -1= ? 

(Complete the solution.) 

3. a. Solve Example 2 by an equation having only one 
unknown. 

b. Which solution do you prefer? 




SIMULTANEOUS LINEAR EQUATIONS 


235 


4. If 2 be added to both numerator and denominator of a 
certain fraction, the resulting fraction equals ^; if 7 be added 
to both terms of the original fraction, the result equals What 
is the fraction ? 


Solution. — 1. 

2. Then, 

3. Also 


Let n = the numerator and 

d = the denominator of the fraction. 

~ = the fraction. 
a 

~ 4 ~ 2 1 

d +2 ” 3 
n±7 _ 1 
2 


( 1 ) 

( 2 ) 


d + 7 

Reduce equations (1) and (2) to the standard from; complete and 
check the solution. 


5. Find two numbers whose sum is 56, and such that \ of 
the greater exceeds f of the smaller by 3. 

6. Solve Problem 5, using only one unknown. 

7. The larger of certain two numbers exceeds twice the 
smaller by 8. Five times the smaller exceeds twice the larger 
by 5. What are the numbers ? 

8. Separate 31 into two parts such that the quotient of the 
larger divided by the smaller is 3, and the remainder is 7. 

9. If 2 be added to the numerator of a certain fraction, the 
resulting fraction equals If 1 be added to the denominator 
of the original fraction, the resulting fraction equals What 
is the fraction? 

10. A’s age now exceeds twice B’s age by 3. Three years 
ago, A was 4 times as old as B was then. What are their ages 
now? 

11. The perimeter of a certain rectangle is 72 inches. The 
width exceeds •§ of the length by 1 inch. What are the length 
and width ? 




236 


ALGEBRA 


12. In an isosceles triangle, two angles always are equal, and 
the sum of all the angles is 180°. In a certain isosceles triangle, 
each of the equal angles is 15° more than one half the third angle 
How large is each angle of the triangle ? 

13. Twice the shorter side of a certain parallelogram exceeds 
the longer side by 6 inches. One half the longer side exceeds 
one seventh the shorter side by 9 inches. How long are the 
sides ? 

14. If three times the smaller of certain two numbers be 
divided by the larger, the quotient is 2 and the remainder is 3. 
The larger exceeds the smaller by What are the numbers ? 

15. A purse contained $5.45 in quarters and dimes. Pay¬ 
ment for a 5j£ purchase was made with a quarter, and dimes 
were received in change. Then the number of dimes was twice 
the number of quarters remaining. How many dimes and 
quarters were there at first ? 

16. Six years ago, A was 12 times as old as B was; three 
years from now, he will be 3 times as old as B is then. What 
are their ages ? 

17. Two angles are complementary. (See page 23.) One 
sixth of the larger exceeds one eighth of the smaller by 4J°. 
How large is each angle ? 

18. The supplement of a certain angle is § of the angle. How 
large is each ? 

19. If the numerator of a certain fraction be doubled and 
the denominator be increased by 5, the resulting fraction equals 
J. If the numerator be decreased by 1, and the denominator 
be increased by 7, the resulting fraction has the value What 
is the given fraction ? 

20. The length of a certain room is 7 feet more than its width. 
If the room is made 1 foot wider and 3 feet longer, the area is 
increased by 66 square feet. What were the original dimensions ? 




SIMULTANEOUS LINEAR EQUATIONS • 237 


Solution . — 1. Let l = the no. of feet in the length; 

2. and w = the no. of feet in the width. 

3. 


4. 


1 = w + 7 


( 1 ) 



Length 

Width 

Area 

Old dimensions 
New dimensions 

l 

l 4" 3 

W 

W + 1 

lw 

(i + SKw + i) 


5. 


(I + 3 )(w + 1) = lw + 66 


( 2 ) 


(Complete the solution, using equations (1) and (2).) 


21. If 5 feet be added to the width and 15 feet to the length 
of a certain lot, the area is increased 1750 sq. ft. If 10 feet be 
subtracted from the length and 15 feet be added to the width, 
the area is increased 750 sq. ft. Find the length and width. 

22. One sum of money is invested at 6%. A second sum, 
which exceeds the first sum by $150, is invested at 7%. The 
interest on the second sum exceeds the interest on the first sum 
by $19. What are the two sums ? 

23. Solve Example 22 by means of only one unknown. 

24. The total value of two investments is $4500. The annual 
interest earned on one is 5%, and on the other is 6%. The 
total interest is $252. What are the two sums ? 

25. The simple interest on $1200 at 7% for a certain number 
of years exceeds the simple interest on $750 at 6% for a second 
period of years by $27. The second period of years exceeds 
the first by 2. How long is each period of years ? 

26. A crew can row 10| miles downstream in one hour and 
5| miles upstream in the same time. What is the rate of the 
current of the river, and of the crew in still water ? 

27. It took a motor boat 1^ hr. to go 20 miles downstream, 
and 2f hr. to return. What was its own rate in still water and 
what was the rate of the current of the river? 










238 


ALGEBRA 


28. Some boys were rowing on a river. It took them only 
three hours to go 20 miles downstream; but it took 10 hours 
to return. Whiat was their own rate of rowing in still water 
and what was the rate of the current ? 

29. On a river whose current is known to be 2-J miles per 
hour, some boys rowed downstream for 2 hours. It took them 
6 hours to return. How far did they go, and what was their 
rate of rowing in still water ? 

30. If a certain lot is made 5 feet longer and 2 feet wider, its 
area is increased 415 square feet. If it is made 3 feet narrower 
and 10 feet longer, its area is decreased 60 square feet. What 
are its dimensions? 

♦ 

158. Relations among digits of a number. Integers are 
written by means of the digits 0, 1, 2, 3, . . .9. 

Thus, 38 is a number of 2 digits; 3 represents 3 X 10 units; 8 repre¬ 
sents 8 units. The sum of the digits is 3 + 8 or 11. 

If t is the tens’ digit, and u is the units’ digit, the number contains 
10 t + u units. The sum of the digits is t + u. 

When the digits are reversed , a new number is formed. Thus, reversing 
the digits of 52 gives 25. 

Notice that 52 = (5 X 10 + 2) units, 
and that 25 = (2 X 10 + 5) units. 

Similarly, if x and y are the tens’ and units’ digits of a number, then 
the number contains 10 x + y units. If the digits are reversed, the 
new tens’ digit is y and the units’ digit is x ; then the new number con¬ 
tains (10 y + x) units. 

EXERCISE 113 

1. How many units are there in the number: 

a. whose tens’ digit is 8 and units’ digit is 5 ? 

h. whose tens’ digit is a and'units’ digit is 6? 

c. whose tens’ digit is b and units’ digit is a ? 

2. The hundreds’, tens’, and units’ digits of a certain number 
are x, y , and z respectively. 

a. How many units does this number contain ? 



SIMULTANEOUS LINEAR EQUATIONS 


239 


b. If the digits are reversed, how many units does the new 
number contain ? 


3. Let t represent the tens’ digit and u the units’ digit of a 
certain number of two digits. 

a. Represent the difference of the digits. (See § 95, page 
123.) 

b. Represent the number itself. 

c. Represent the quotient of the tens’ divided by the units’ 
digit. 

d. Represent the number with digits in the reverse order. 

e. Represent this last number divided by the sum of the 
digits of the given number. 

4. The sum of the digits of a certain two-digit number is 
11. If the digits be reversed, and the new number be divided 
by the old, the quotient is 2 and the remainder is 7. What is 
the number? 


Solution. — 1. Let t = the tens’ digit of the number, 
and u = the units’ digit of the number. 


t + u = 11 


( 1 ) 



TeN8’ 

Digit 

Units’ 

Digit 

Number Itself 

Given number 

t 

u 

10 t + u 

New number 

u 

t 

10 u+t 


.*. (10 u + t) = 2(10 t + u) + 7. (2) 


(Complete the solution, using equations (1) and (2). Check the 
solution, by reference to the statements in the given problem.) 


6. The tens’ digit of a certain two-digit number exceeds 
its units’ digit by 5. If the digits be reversed, the new number 
is one fourth of the result obtained by adding 3 to the given 
number. What is the number ? 

6. The tens’ digit of a certain two-digit number exceeds 
twice the units’ digit by 1. If the digits be reversed, the sum 















240 


ALGEBRA 


of the new number and the original number is 143. What is 
the number? 

7. The sum of the two digits of a certain two-digit number 
is 9. If the digits be reversed, the resulting number is 9 less 
than 3 times the original number. What is the number? 

8. A certain number consists of two digits. If the digits 
be reversed, the sum of twice the original number and the new 
number is 168. Also the original number exceeds 4 times the 
sum of its digits by 3. What is the number? 

9. A certain number consists of three digits of which the 
units’ digit is 5. The sum of the digits is 12. If the hundreds’ 
and tens’ digits are reversed, the new number is 90 less than 
the original number. What is the number? 

10. In a certain number consisting of three digits, the tens’ 
digit is 6. The hundreds’ digit is 1 less than twice the units’ 
digit. If the hundreds’ and units’ digits be interchanged, the 
new number exceeds one half the original number by 85. What 
is the number? 

EXERCISE 114 

Review Examples 

1. Divide 4 m 3 4 * 6 — 9 m 2 -f 6 m - 1 by 2 m 2 + 3 m — 1. 
Check, letting m = 2. 

3 c 2 — 6 c 

2. Reduce to lowest terms —--• 

G c 2 - 24 

3. Solve for k : — - — (fc - 11) = § (k - 25) + 34. 

5 14 7 

4. a. Solve T = 2 ttR(R + H) for H in terms of T, 7 r, 
and R. 

b. In your result, let 7 r = 3.14, R = 10, T — 794.42. Find 
H to two decimal places. 

6. a. The three digits of a number are a, b , and c. Repre¬ 
sent the number formed by reversing the digits. 



SIMULTANEOUS LINEAR EQUATIONS 


241 


b. A man earns e dollars a month, and spends s dollars a 
month. How much will he have saved in 3 years ? 

c. John is y years old now. Write the equation expressing 
the fact that 4 times his age 3 years ago is twice his age 5 years 
from now. 


6. a. Simplify — a -— ^ - 


+ 


u + 3 


a 2 — a — 6 a 2 — 4 a + 3 a 2 + a — 2 


b. Check the solution by letting a = — 1. 


7. Simplify 


4 r 2 — 12 rs -f 9 s 2 4 r 2 — 4 s 2 3 rs — 3 


r 2 + 4 rs + 4 s 2 4 r 2 — 9 s 2 5 r 2 + 10 


rs 


8. Solve and check 6 f + 5 + —L_ _ 3 * 

2 a 2 - 2 z z 2 - 1 x 2 - 1 


9. Solve and check 


x — b 


x + 6 


4 a 2 


x — 2 a +2a a: 2 - 4 a 2 


SUPPLEMENTARY TOPICS 
159. Literal simultaneous equations. 

Example. — Solve for x and y the system: 




f ax + 

by 

= 

c 

(1) 



\rx + 

y 

= 

t 

(2) 

Solution. — a. 

Ehminate y. 





1. 

M s (1) 

asx + bsy 

— 

cs. 


(3) 

2. 

M b (2) 

brx + bsy 

= • 

bt. 


(4) 

3. 

(3) - (4) 

asx — brx 

= 

cs 

— bt. 

(5) 

4. 

Factoring, 

x(as — br) 

= 

cs 

- bt. 



5. D(as—6r) 


cs — bt 


as — hr 

b. Next, going back to (1) and (2), eliminate x. 

arx + bry = cr. 
arx + asy = at. 
bry — asy = cr — at. 
y(br — as) = cr — at. 
cr — at at — cr 


6. M r (1) 

7. Ma (2) 

8. (6) - (7) 

9. Factoring, 

10. D(ftr-os) 


V = 


or y = 


br — as 

11. .*. The common solution is x = 


as — br 
cs — bt 


at — cr 


—rr > y = u 
as — br as — br 


( 6 ) 

(7) 

;(8) 



















242 


ALGEBRA 


Note 1. — Step 3 can be omitted if you notice that asx and brx are 
like terms, and that their difference can be obtained by multiplying 
their common factor x by the difference of its coefficients; namely, 
(as — br). 

Note 2. — Remember that this result is true for all values of a, b, 
c, r, s, and t which do not make any denominators equal to zero. 


EXERCISE 115 


Solve the following systems for x and y : 


\2x -Zy = a 
\3x + 4y = b 

j cx cy =2 a 
\ cx — cy = 2 b 

f 4 x + ay = b 
\ 5 x + by = a 

I bx + y = b 
[ ax + 2 y = a 


g \2x-\-y=3a + b 
\x + 2y = 3a + 2b 

9. |2* + 3j/=5a + 6 
\3x +2 y.= 5 a - b 

1Q ( mx + y = m 2 + n 
\ x + my = m + inn 

n l rx + ry = 2 r 2 
\ sx — sy = 2 s 2 


| ax + by = 1 
1 x + y = 1 

f ax — by = 1 
\ mx + ny = 1 


| bx -f ay = a -f b 
\ abx + aby = a 2 -f b 2 

( ax = by 

abx -f- aby = (a + b) 2 


7. 


f ax — by = c 
\dx - ey = f 


\ bx — ay = ab 
\ 3 ax + 3 by = 2 a 2 — b 2 


15. a. Find two numbers whose sum is m and difference is n. 
b. Using the results as formulae, find the numbers whose sum 
is 33 and difference is 7. 


16. a. Separate n into two parts such that the larger will 
exceed twice the smaller by m. 

b. Using these results as formulae, find the two parts when 
n is 50 and m is 8. 


SIMULTANEOUS LINEAR EQUATIONS 243 


17. a. Separate a into two parts such that when the larger 
is divided by the smaller the quotient is b and the remainder 
is c. 

b. What are these two parts when a = 75, b = 2, and c — 3 ? 

c. What are these two parts when a = 100, b = 4, c = 5 ? 

18. a. A is now r times as old as B. s years ago A was t 
times as old as B. What are their present ages ? 

b. What are these ages when r = 2, 5 = 15, £ = 5? 

c. These formulae will always give some sort of results. The 
results may be impossible because the conditions are impossible. 
Thus, let r = 5, s = 10, t = 3. 

19. a. A sum of money amounting to m dollars consists of 
quarters and dimes alone. The number of quarters is n more 
than the number of dimes. How many of each coin are there? 

b. What are the results when m = 5 and n = 6 ? 

c. What are the results when m = 5.6 and n — 7 ? 

d. What are the results when m = 5 and n = 3 ? 

20. a. If m be added to the numerator of a certain fraction, 
the resulting fraction has the value c. If n be subtracted from 
the numerator of the fraction, the resulting fraction has the 
value d. What is the fraction ? 

b. What is this fraction when m = 4, c = 2, n = 2, and 
d = 1? 

21. a. The length of a certain rectangle exceeds its width by 
c feet. The perimeter is p feet. Find its length and width. 

b. Using the results of part a as formulae, find the length 
and width when c is 7 feet and p is 54 feet. 

c. Repeat part b when c is 8 feet and p is 90 feet. 

22. a. There are two angles which are complementary. One 
exceeds the other by n degrees. How large is each ? 

b. Using the results of part a as formulae, find the two 
angles when n is 15°. 

c. Repeat part b when n is 23°. 


244 


ALGEBRA 


160. Fractional equations linear in - and - • 

x y 


Example. — Solve the system 


Solution. — 1. Mg (1) 

2. M 3 (2) 

3. Adding, 


50 _ 45 ; 
x y 
24 ,45. 


x y 
74 _ 
x 

74 = 37 x, or * = 2. 
9 . 


12-2-8. 

(1) 

x y 



(2) 

x y 


■■ 40. 

(3) 

= -3. 

(4) 

37. 

(5) 


5. Substitute 2 for x in (1) 5 — = 8 . 

y 

6. Solving for y, y = — 3. 


Solve 

1 


EXERCISE 116 

the following systems of equations : 


' 4 + ? = 5 
x y 


12 _6 = _i 

x 
9 


y 

10 


■ + — = — 8 

y 

' 4 +? = 7 

X 


y 

10 5 


15 

2 


r o 17 

2y -* = i 

3 !/ + - = — 1 

L X 


' 5 _ 8 = j 

x y 

3 + ? = - 13 
l x y 


_2 

A 

7 


15 . 

B 


,5 = _ 3 
A B 4 


f JL. 

! 2 m 


13 

4 


U + 2 = 25 
[m n 6 
Hint. — M (1) by 4. 


8 . 


[2 + !--io 

! a b 


t _1_1_ m _ 35 

[ 3 a 2 b 12 
— M (2) by 12. 





SIMULTANEOUS LINEAR 

EQUATIONS 

245 

31 _ _ 1 

2r 3s 2 



f± + A=- 

Zc id 

11 

10. 


2 

5 3_ = _ 17 

r 2s 6 


1 2 

9 



6c 5 d 

5 

Work problems and problems 

about reciprocals 



Note. — For work problems, review Examples 1-4, page 192. 

The reciprocal of 3 is 1 -r 3 or £; of $ is 1 -s- f or f; of x is - • 

x 

11. The sum of the reciprocals of certain tw<$ numbers is J. 
Twice the reciprocal of the smaller, diminished by the reciprocal 
of the larger, also is What are the numbers ? 

Suggestion. — If l = the larger, then 1 is its reciprocal. 

12. There are two unequal numbers such that the reciprocal 
of the smaller, diminished by the reciprocal of the larger, is f. 
Twice the reciprocal of the smaller plus five times the recip¬ 
rocal of the larger equals 6. What are the numbers ? 

13. A and B together do a certain piece of work in 10 days. 
They can do it also if A works 12 days and B works 6 days. 
How long would it take each to do it alone ? 

Suggestion. — If A can do it in a days, how much can he do in 1 day? 
in 10 days? etc. 

14. A and B together can do f of a certain piece of work in 
3 days. They can do all of it if A works 9 days alone and B 
works 2 days alone. How long would it take each of them to 
do all of it alone ? 

15. A workman and one helper did a certain piece of work 
in 6 days. On another occasion, the workman and two helpers 
did the same work in 4 days. If the two helpers worked at 
the same rate, how long would it take the workman alone or 
each of the helpers alone to do the work ? 

161. Equations containing three variables. If there are 
three variables, there must be three independent equations 
in the system. 







246 


ALGEBRA 


EXERCISE 117 

Solve the following systems of equations. 


{2 x — y + z = 5 (1) 

1. \3x + 2y + 3z = 7 (2) 

l4*-3y-5a--3 (3) 

Solution. — a. Eliminate z from (1) and (2). 

1. M 3 (1) 6 x - 3 y + 3 2 = 15 . (4) 

2. Subtract (2) from (4) | 3 x — 5 y = 8 | (5) 

b. Eliminate aifrom (1) and (3). 

1. M 5 (1) 10 x — 5 ?/ + 5 z = 25. (6) 

2. Add (3) and (6) 114 x - 8 y = 221 (7) 


c. Complete the solution by solving the system consisting of equa¬ 
tions (5) and (7), thus finding x and y. 

d. Substitute in (1) the values found for x and y. Solve for z. 

e. Check by substituting in the three original equations. 


[2x — y + z = 3 

2. <j3x-f2y-j2=4 

[4* + 3 y — 2 z = 4 

f 3x + 4y + z = -1 

3. (2x—y+2z = — 1 
[x + 3y - z = 2 

(4x — y + 2 z = 1 

4. 3z + 2?/+3z=4 
[ 5 x — 7 y = -3 

{ 3 w -f 2 r = 9 
6. | 2z - r = 11 
I 5 w — 6 z = 1 


( 6 ^ + 3 t + 2 « = 1 

6. |3s-4tf=4 
[ 5 s — t = 14 

[2 A +3 B -C = 0 

7. 4J-6R+C=2 
[ A - B + C - 2^ 

f8m — 3n — 12c= —2 

8. |3m—w — 6c=—^ 

[ 5 m + 9 n = \ 

\2x — 2 y + 3 z = 

9. j 3 x — 2 2 = f 

[ 4 y + 10 z = 5 


10. In a certain triangle, twice angle A exceeds angle B by 
30°, and twice angle A exceeds angle C by 40°. Find the angles 
of the triangle. (See page 25, Example 22.) 

11. The perimeter of a certain triangle is 165 inches. The 
sum of the first and second sides equals twice the third side; 






SIMULTANEOUS LINEAR EQUATIONS 247 


twice the first side exceeds the second side by 10 inches. How 
long are the sides of the triangle ? 

12. A certain publishing company offered three magazines 
called A, B, and C at the following combination rates : 

For A and B together $3.50 
For A and C together $5.50 
For B and C together $5.00 

What is the price at which each magazine is figured in these 
combinations ? 

13. The sum of the three digits of a certain number is 14. 
The units’ digit exceeds the sum of its hundreds’ and tens’ digit 
by 2. The tens’ digit is one half the units’ digit. What is the 
number ? 

14. The sum of the three digits of a certain number is 16. 
If the digits be reversed (see page 238), the new number ex¬ 
ceeds the old number by 99. The sum of the tens’ digit and 
units’ digit equals 3 times the hundreds’ digit. What is the 
number ? 

16. The sum of the three digits of a certain number is 7. If 
the tens’ and hundreds’ digits he interchanged, and the new num¬ 
ber be divided by the old, the quotient is 2 and the remainder 
is 20. The tens’ digit exceeds twice the hundreds’ digit by 1. 
What is the number ? 

16. A and C together can do a piece of work in 4 days. If 
A and B work together for 3 days, then B can complete the work 
in 1 more day. If B and C work together for 4 days, C can 
complete the work in 2 days more. How long would it take 
each to do it alone ? 


XIII. SQUARE ROOT AND QUADRATIC SURDS 


162. The square root of a perfect square monomial was 
found by inspection in § 91, page 119. Review this section. 

163. Every number has two square roots. They have equal 

absolute values but opposite signs. 

Thus, V«9 4 6 2 = + 3 a 2 b, because (+ 3 a 2 b) 2 = 9 a 4 b 2 . 

Also, V9 a 4 6 2 = — 3 a?b, because (— 3 a 2 b) 2 = 9 a 4 #*. 

•These two roots are written together by means of the double 
sign ±, read plus or minus. 

Thus, V9 a A b 2 = ± 3 o 2 b ; read plus or minus 3 a 2 b. 

Of these two square roots, the positive one is called the prin¬ 
cipal square root. 

When the square root is mentioned, usually the principal 
square root is meant. 

164. The square root of a large number often may be obtained 
by inspection, after factoring the number. 

Thus, V1764 a 4 = V4 • 441 a 4 = V 4 • 9 • 49 a 4 = 2 • 3 • 7 • a 2 = 42 a 2 . 
This solution uses the following fundamental principle: 

Rule. — The square root of the product of two or more num¬ 
bers equals the product of their square roots. 

Thus, V4 -25 = 2 • 5 = 10; Vab = \4 • Vb. 


EXERCISE 118 
Find the square roots of: 


1 . 

36 m 4 

5. 196 rh 4 

9. 

1089 a: 4 

2. 

64 a 6 i 2 

6. 225 & 

10. 

1764 y e 

3. 

121 mV 

7. 900 a 2 

11. 

1936 2 V 

4. 

169 a: 6 

8. 576 c 4 

12. 

2304 mV 


248 











SQUARE ROOT AND QUADRATIC SURDS 249 


13. Find the square roots of 9 a 2 — 12 ax + 4 x 2 . 

Solution. — 1. V9 a 2 — 12 ax + 4 x 2 = V(3 a — 2 x){% a — 2 x). 


V9 a 2 - 12 ax + 4 x 2 = V(3 a - 2 z) 2 
= =*= (3 a — 2 x). 


2 . 


14. 25 a 2 - 10 a + 1 

15. c 2 — 16 c -f 64 


17. 4 a 2 — 20 a -f 25 

18. 9 a 2 - 30 ab + 25 b 2 


16. m 2 — 20 mn -j- 100 n 2 19. 16 c 2 — 24 cd + 9 d 2 

Example 20. — Make a perfect square trinomial of 
25 x 2 — 30 xy -f* ( ? ) by supplying the missing term; then find 
the principal square root of the result. 

Solution. — 1. In the expressions of Examples 13-19 observe: 

Two terms (the first and third) are perfect squares, preceded by plus 
signs; the remaining term is twice the product of their square roots. 

2. In this example, V 25 x 2 = 5 x. 

3. — 30 xy must be 2 • (5 x) • some number. 

.*. that number must be 3 y, since 30 xy -f- 10 x = 3 y. 

4. The remaining term must be (3 y) 2 or 9 y 2 . 

5. The perfect square trinomial is 25 x 2 — 30 xy + 9 y 2 

6. and V25 x 2 — 30 xy + 9 y 2 — (5 x — 3 y). 

Make perfect square trinomials of the following; then give 
the principal square root of each : 


21 . a 2 + 6 a + (?) 


22 . r 2 — 12 r + (?) 


28. z 2 + |z + (?) 

29. * 2 -§* + (?) 


23. t 2 - 18 tw + (?) 


24. m 4 — 24 m 2 x + (?) 


30. 36 z 2 - 12 * + (?) 

31. 49 m 4 — 14 m 2 n + (?) 


25. x 2 - (?) + 36 y 2 

26. a^ + 3a; + (?) 

27. 2/ 2 -5y + (?) 


32. 9 a 2 - (?) + 36 b 2 

33. 49 a; 2 - 42 a: + (?) 

34. 36 a 2 - 60 ab + (?) 







250 


ALGEBRA 


165. The square root of a perfect square polynomial can be 
found by a long division process. 

Example 1. — Find the square root of 9 x 2 + 25 y 2 — 30 xy. 
Solut ion. — 1. Arrange the polynomial in descending powers of x. 

2. Vq x 2 = 3 x. Write 3 x in the root. Root 3jr_ —5 y 

3. Square 


9 x 2 . 


9 x 2 — 30 xy + 25 y 2 
9x 2 


6 x 


- 5 y 
6 x — 5 y 


-30 xy + 25 y 2 


— 30 xy + 25 y 2 


5 y), which is 


3 x, getting 
Write it below 9 x 2 . 

4. Subtract, obtaining the first 

remainder. 

5. Trial divisor, 2 • 3 x = 6 x. 

— 30 xy -*• 6 x = —5 y. 

Write — 5 y in the root and 
add — 5 y to 6 x, forming 
the complete divisor. 

6. Multiply the complete divisor 

by — 5 y. 

7. Subtract. In this case, no remainder. 

8. The square roots are =•= (3 x — 5 y) or + (3 x 
3 x — by, and — (3 x — 5 y), which is — 3 x + 5 y. 

Note. — All the explanation of this solution should be carried along 
mentally. Only the solution is written. 

Example 2. — Find the principal square root of 
20 a* — 70 x + 4 x 4 + 49 - 3 x 2 . 

Soluti on. — 1. Arrange the polynomial in descending powers of x. 

2. V4 x 4 = 2 x 2 . Write it 2 x 2 + 5 * - 7 

in root. 

3. (2 x 2 ) 2 = 4 x 4 . Write it. 

Subtract. 

4. Trial divisor, 2 • 2 x 2 . 

5. 20 z 3 4 x 2 = 5 x. 4 x 2 

Write it in root. _ -f 5 x 

6. Add + 5 x to 4 x 2 . 


4 x 4 + 20 x 3 - 3 x 2 - 70 x + 49 


4 x 4 


+ 20x 3 — 3x 2 - 70x + 49 


4 x 2 + 5 x 

7. Multiply 4 x 2 + 5 x by + 5 x. 

Subtract. 4 x 2 + 10 x 

8. Trial divisor, 

2(2 x 2 + 5 x). 

9. - 28 x 2 -r4x 2 = - 7. 

Write it in root. 

10. Add - 7 to 4 x 2 + 10 x. 

11. Multiply 4 x 2 + 10 x — 7 

by — 7. Subtract. 


20 x 3 + 25; 


- 7 


4 x 2 + 10 x - 7 


- 28 x 2 - 70 x + 49 


- 28 x 2 - 70 x + 49 

















SQUARE ROOT AND QUADRATIC SURDS 251 


EXERCISE 119 


Find the square root of: 

1. 25 x 2 + 40 xy + 16 y 2 

2. 9 a 2 - 30 ab 4- 25 b 2 

3. 36 c 2 - 60 cd 4- 25 d 2 

4. 49 m 4 — 28 m 2 n 4-4 n 2 
6 . 25 r 2 - 70 rs -f 49 s 2 

11. 4 c 4 - 12 cH 4- 


6. 4 z 4 4- 4 z 3 4- 5 z 2 4- 2 X + 1 

7. 4 z 4 - 4^4-5 z 2 - 2 z 4- 1 

8. 9ar 4 + 6z 3 + 13* 2 + 4a;4-4 

9. 9 a 4 4- 12 a 3 - 2 a 2 - 4 a 4- 1 
10 . 2/ 4 — 6 2Z 3 4- 13 y 2 - 12 y + 4 
c 2 d? - 12 cd? 4- 4 d 4 


12. 9 m 4 — 24 m 3 n 4- 10 m 2 n 2 4~ 8 mn 3 4* n 4 

13. 1 - 24*— z 2 4- 2 a: 3 4- a: 4 

14. a 2 4- 4- c 2 4- 2 a6 4- 2 ac 4- 2 be 

15. 4 a 2 4- 9 6 2 -f c 2 4- 12 a6 — 4 ac — 6 be 

16. 9 a 2 — 24 ab 4- 30 ac 4- 16 b 2 - 40 6c -f 25 c 2 

17. 4 a; 2 — 12 xy — 16 xz 4- 9 y 2 4- 24 yz 4- 16 z 2 

18. x 4 -\-2x 3 +2x 2 + x-\~i 

19. x 4 - x 2 y 4- f X 2 z 4- i y 2 - i yz + £ z 2 

20 . -^a 4 — 2 a 2 6 4- a 2 c 4- 4 6 2 — 4 6c 4- c 2 


166. The square root of an arithmetical number. 

Any number having one or two figures to the left of the 
decimal point (like 1, or 99.5) equals or exceeds 1 and is less 
than 100; therefore its square root, equal to or exceeding 1 and 
less than 10, must have one figure to the left of the decimal point. 

Any number having three or four figures to the left of the 
decimal point (like 100.25 or 9999.7) equals or exceeds 100 and 
is less than 10,000. Therefore its square root, equal to or ex¬ 
ceeding 10 and less than 100, must have two figures to the left 
of the decimal point. 

Hence, if the given number is divided into groups of two 
figures each, starting at the decimal point, for each group in 
the number there will be one digit in the square root. The 
groups are called periods. 



252 


ALGEBRA 


Thus, 2345 becomes 23 45. It has two periods. Its square root 
has two figures to the left of the decimal point, a tens’ figure and a 
units’ figure. 34,038 becomes 3 40 38. It has three periods. Its 
square root has three figures to the left of the decimal point. The left¬ 
most period may contain only one figure, as in this case. 

A decimal is divided in the same manner, counting in both directions 
from the decimal point. Thus, 34256.895 becomes 3 42 56.89 50. 
The square root of this number has three figures to the left of the 
decimal point and two to the right. 

The square root is found as follows: 

Example 1. — Find the square root of 4624. 

Solution .— 1. 4624 must be the square of a number more than 60 
and less than 70, since 60 2 = 3600 and 70 2 = 4900. 

That is, the square root is 60 + some number less than 10. 





a 


b 




60 + 8. 


6 8. 


60 2 = 3600. Subtract from 4624. 


46 24. 


46 24. 

2 . 

Trial divisor, 2 X 60 = 120. 


36 00 


36 

3. 

1024 -j- 120 = 8 + . Add 8 to 120. 

120 

10 24 

' 120 

10 24 


Add 8 to 60 in the root. 

8 


8 


4. 

Multiply 128 by 8. Subtract. 

128 

10 24 

128 

10 24 


Note. — The solution is always written in the form 6. The canceled 
zeros in solution a are omitted in solution b. 


Rule. — To find the square root of a number: 

1. Separate the number into groups of two figures each, 
starting at the decimal point, and forming the groups each way 
from the decimal point. 

(Each group is called a period. The left-most period may contain 
only one figure. Thus, 62753.489 becomes 6 27 53.48 90.) 

2. Find the largest square number not more than the left-most 
period. Write its square root as the first figure of the square 
root. Subtract the square number itself from the first period. 

3. To the remainder, annex (bring down) the next period. 

4. Form the trial divisor by doubling the root already found, 
and annexing a zero. 










SQUARE ROOT AND QUADRATIC SURDS 253 


6 . Divide the remainder formed in Step 3 by the trial divisor. 
Annex the quotient to the root already found, and add it to the 
trial divisor to form the complete divisor. 

6 . Multiply the complete divisor by the new figure of the root. 
Subtract the result from the remainder formed in Step 3. 

7. If other periods remain, repeat Steps 4, 6, and 6 until there 
is not a remainder, or until the desired number of decimal 
places for the root have been obtained. 


Note. — In Step 6, sometimes the product is greater than 
mainder. In such cases, the last figure obtained for the n 
large. Substitute for it the next smaller integer. 

Sometimes the next figure of the root must be madey^ro. In this 
case, bring down the next period and repeat Steps 4, 5, (n i 6 

Example 2. — Find the square root of 5207.0656 

Solution. — 1. The largest square number in 52 
is 49. Write V49 or 7 in the root. Subtract. An¬ 
nex 07. 



52 07.06 56 
49 


2. Trial divisor is 2 X 7. Annex 0. 307 -5- 140 

140 

3 07 

= 2 + . Place 2 in the root. Add 2 to 140. Mul¬ 

2 



tiply 142 by 2. Subtract. Annex 06. 

142 

2 84 

3. Trial divisor is 2 X 72. Annex 0. 2306 + 

1440 

23 06 

1440 = 1 + . Place 1 in the root. Add 1 to 1440. 

1 



Multiply 1441 by 1. Subtract. Annex 56. 

1441 


14 41 

4. Trial divisor is 2 X 721. Annex 0. 86556 -f- 

14420 

8 65 56 

14420 = 6 + . Place 6 in the root. Add 6 to 14420. 


6 


Multiply 14426 by 6. No remainder. 

14426 

8 65 56 


The square root is 72.16. 


EXERCISE 120 

Find the square root of: 


1 . 

1296 

6 . 

17,689 

11 . 

164,025 

16. 

9.4249 

2 . 

1849 

7. 

33,489 

12 . 

1474.56 

17. 

761.76 

3. 

3721 

8 . 

82,369 

13. 

19.1844 

18. 

5913.61 

4. 

7396 

9. 

96,100 

14. 

4160.25 

19. 

8.5264 

5. 

8836 

10 . 

43,264 

15. 

62.8849 

20 . 

64.6416 













254 


ALGEBRA 


167. The approximate square root of a number which is not 
a perfect square is often desired. 

Example. — Find V2 to three decimal places. 


Solution. — 


1. 4 1 4 


2.00 00 00 
1 

1.00 

96 


280 

4 00 

1 


281 

2 81 

2820 

1 19 00 

4 


2824 

1 12 96 


6 04 


Therefore the square root of 2 to three decimal places is 1.414. The 
root, correct to hundredths, is 1.41, since the figure in the third decimal 
place is less than 5. If it were more than 5, the figure in the hundredths 
place would be increased by 1. Thus, 3.146, correct to hundredths, is 3.15. 


EXERCISE 121 

Find the approximate square root to three decimal places: 

1. 3 3. 6 6. 10 7. 13 9. 15 11. 19 

2. 5 4. 7 6. 11 8. 14 10. 17 12. 21 


168. Table of square roots. In the remainder of your courses 
in mathematics, you will need frequently the square roots of 
certain numbers. It is not economical to compute the same 
roots repeatedly. Therefore keep in a notebook any square 
roots which you compute. Begin now by writing the numbers 
from 1 to 50 in a column. Opposite each, in a second column, 
write the principal square root of the number to three decimal 
places. 








SQUARE ROOT AND QUADRATIC SURDS 255 

After you have completed Exercise 121, you will have 12 of 
these square roots. Some others you can write down by in¬ 
spection. Others can be got as follows. 

Example. —V^8 = V^4 • 2 = 2V2 = 2 X (1.414) = 2.828. 


EXERCISE 122 

Find each of the following, using any roots which you have 
already computed. 

1. Vl2 3. V20 6. V27 7. V45 9. V / 40 

2 . Vl8 4 . V24 6. V 28 8. V32 10 . V 44 


11 . Complete the table of square roots to 50: 

a. getting as many as possible by inspection; 

b. getting others as in Examples 1-10 above; 

c. getting the rest by the process taught in § 166. 


169. The square root of a fraction may be obtained in three 


ways. 

Example 1. — Find Vf to three decimal places. 

Method a. % = 1.5. The square root can be found as in § 166. 
It cannot be got mentally. 


Method b.-yj: |=^=LZ32 =? 


long division. 
Method c. 


'2 V2 l- 414 


The result must be found by 


.13 _ /6 _ V6 = 2.449 = j 224 

Va-A/i-vf 2 


In methods b and c, the square roots are taken from your 
table of roots. If you do not have the necessary roots, then 
two roots must be computed in method b, and only one in 
method c. 

Method c is clearly the one to use. Observe that the fraction 
f is changed to an equivalent fraction (see page 169), whose 
denominator is the smallest possible perfect square. 


256 


ALGEBRA 


Rule. — To find the square root of a fraction: 

1. Change the fraction to an equivalent fraction whose de¬ 
nominator is the smallest possible perfect square. 

2. The square root of the fraction equals the square root of 
its numerator divided by the square root of its denominator. 

3. Express the result of Step 2 in simplest radical form, and 
also in decimal form, to three places. 

Example 2. — Find Vf • 


Solution. — 1. The smallest perfect square into which 8 can be 
changed is 16. Multiply both terms of the fraction by 2. 


2 . 



2.449 

4 


.612. 


EXERCISE 123 


Find the square root of: 


1-1 

4- 1 

7. * 

10. i 

13. * 

2- f 

e. t 

8- § 

11. f 

14- A 

3- t 

6- t 

9- I 

12. * 

is- A 


170. A quadratic surd is the indicated square root of a num¬ 
ber which is not a perfect square; as V3. The square root 
symbol indicates the principal square root. 

A surd is in its simplest form when the number under the 
radical sign is an integer which does not have any perfect square 
factors. 

Thus, V24, simplified, becomes V4 • 6 or 2 V6. 

Vf, simplified, becomes or ^VlO. 


171. A surd expression is an expression involving one or more 
surds. In this chapter, only quadratic surd expressions are 
considered. 

A surd expression is simplified by first simplifying the surds, 
and then performing all other indicated operations. 


SQUARE ROOT AND QUADRATIC SURDS 257 


Example. — Simplify f — Vf. 

solution. —i. i~Vf = i “V! 

o 3 vli 

2 ' 7 7 

, 3 — vTi 

3. - ? 

Find the value of the expression to three decimal places. 

Solution. - 3 ~ = 3 ~ 3.741 = ^741 _ _ m 

7 7 7 

EXERCISE 124 

a. Simplify the following expressions; b. express the results 
to three decimal places : 


1. 6 + Vl2 

6. § + 

11. 


2. - 5 + Vl8 

7. f - Vj 

12. 

-1 + 

10 

1 

CO 

8. | + vT 

13. 

-t-v§ 

4. - 3 - 

9. f — vT 

14. 

15 -v/ 5 ' 

+ 6 V 

6. + 7 + V 24 

10. f + v| 

15. 

— 5-4- V 5 , 

^ T v TS- 


SUPPLEMENTARY TOPICS 

172. Further addition and subtraction of surds. 

Example 1. — Find V20 + V45. 

Solution. — 1. V 20 + V45 = V4T5 + ViTlj 

2. = 2V5 + 3V5 

3 . = 5V5. 

This is the sum in its simplest radical form. In this form, its value 
to any desired number of decimal places is found with a minimum of 
computation. 

5 V 5 = 5 X 2.236 = 11.180. 

The surds expressions 2 V 5 and 3Vo are called similar surds, be¬ 
cause the surd factor of each is the same. Only similar surds can be 
combined. 






258 


ALGEBRA 


Example 2. — Find the simplest radical form and also the 
decimal value to three places of V§ -f V|. 

Solution. — 1. >/§ + V| = VlT + Vf 

o _ 3V2 , V2 

4 2 

o _ 3 V 2 + 2 V 2 _ 5 V 2 

3 . 4 ^ or — 

4 5V2 = 5 X 1.414 = 7,07 = 17Q7 

4 4 4 


EXERCISE 125 

Find the simplest radical form and also the decimal value to 
three places of the following expressions : 


1. Vl2 + V27 

2 . V75 - V48 

3. Vl8 + V32 

4. V45 - V5 
6 . 3V6 - V24 

6. V28 - 3V7 - V63 

7 . V 50 - Vl8 + V& 

8 . V80 - V45 + V5 

9. V48 - 2V3 + V300 
10. V200 - V98 - V 72 


11 . V\ + Vs 

12. Vf + v\ 

13. Vf - V> 

14. Vf + V 40 

15- V% - V* 

16- + VI - v\ 

17. V| - Vjr + V* 

18 . V* + V 90 - V4^ 

19. V i + Vi-Vl 

20. V\ + V20 - V±£ 


173. Multiplication of quadratic surds. 

The fact that Va6 = Va • V& has been used repeatedly. 
Thus, 

V4^9 = V 4 • V 9 = 2-3 = 6; also V 4 T 3 = VI . V 3 = 2 V 3 . 
Since Va& = Va • Vfe, then also Va • V& = Va6. 

Thus, V2 • V 3 must equal V 2 • 3, or Vq. 





SQUARE ROOT AND QUADRATIC SURDS 259 

Rule. — The product of the square roots of two numbers 
equals the square root of the product of the numbers. 

Example. — Find the simplest radical form and also the 
decimal value to three places of 2V3 x 5V2. 

Solution. — 1. 2V3 X 5V2 = 10V3 • 2 = lOV^. 

2. 10^6 = 10 X 2.449 = 24.49. 

When doing such multiplication, keep the product under the radical 
sign in factored form. 

Thus, Vl8 • Vl2 = Vl8 • 12 = V9 • 2 • 4 • 3 = 3 • 2^6, or 6^6. 


EXERCISE 126 

Find the simplest radical form and also the decimal value to 
three places of: 

1. V2 • V8 4 . VQ • Vl2 7 . 3V2 • 4V5 10 . (3V3) 2 

2 . VS • Vl2 5 . V7 • V21 8 . 2v / 6 • 5V3 11 . (2V2) 2 

3 . V5 • V15 6 . Vn • V22 9 . 6V20 • 3Vld 12 . (iV2) 2 

13 . Multiply 3 + 5 V 2 by 3 + 5 V 2 . 

Solution. — 3 + 5V2 

3 + 5V2 

9 + 15 V2 

15V2 + 25V4 

9 + 30V2 + 50 or 59 + 30^2 
The decimal value of the product is 59 + 30 X 1.414, or 101.42. 


Find: 

14 . (2 - V3)(2 - V3) 

16 . (5 + V6)(5 + V6) 

16 . (- 3 + V2)(- 3 + V2 ) 

17 . (9 - a/5) (9 - a/5) 

18 . (-7+a/II)(-7 +VTT) 

19. (10 - a/6)(10 + a/6) 


20. (3 - 2a/2)(3 - 2 V 2 ) 

21. (5 + 3V7)(5 + 3a/7) 

22. ( - 4 + 5V3) 2 

23 . (6 - 2V7f 

24 . (-3 +5a/6) 2 
26 . (11 - 3A/5) 2 






XIV. QUADRATIC EQUATIONS 

174 . A pure quadratic equation is an equation in one un¬ 
known in which the unknown does not appear in the denomina¬ 
tor of a fraction, and in which only the second power of the 
unknown does appear. 

Thus, 5 x 2 — 37 = 0 is a pure quadratic equation. 

Example. — How long must the side of a square field be in 
order that the area of the field shall be one acre (43,560 sq. ft.) ? 

Solution. — 1. Let s = the no. of ft. in one side of the square. 

2 . .*. s 2 = the area of the square. 

3. /. s 2 = 43,560. 

4. Take the square roo t of bo th members of the equation. 

.*. s = V43,560- 

5. .'. 8 = =*= 208.7 ft. 

6 . Since s is the length of a side of the field, only the positive root 
has meaning. That is, the side must be 208.7 + ft. 

175. A pure quadratic equation has two roots. 

Example. — x 2 = 16. 

Solution. — When we take the square root of both sides, really we 
get =*= x = =*= 4 This means: 

(1) + x = + 4 (3) - x = + 4 

(2) + x = - 4 (4) - x = - 4 

However, if both members of equation (3) are multiplied by — 1, 
we get + x = — 4, which is equation (2); and if both members of 
equation (4) are multiplied by — 1, we get + x = +4, which is 
equation (1). 

That is, we actually get only two equations, and not four. 

We get these two equations, moreover, if we use only the positive 
root on the left side and the two roots on the right side; that is, + x 
= ±4. 


260 




QUADRATIC EQUATIONS 


261 


Rule. — To solve a pure quadratic equation : 

1. Simplify the equation until it takes the form x 2 = a number. 

2. Take the square root of both members of the equation, 
placing the + sign in front of x, and the double sign (±) before 
the square root of the right member. 


Example. — Solve the equation — 

3 m 12 m 


Solution. — 1 . 

2. M 12 m 

3. Simplifying, 

4. M 7 

5. m 

6 . 

7. 


2 m i _3 _ m , 12 
3 m 12 m 
8 m 2 + 36 = m 2 + 144. 

7m 2 = 108. 
m 2 = 

= ± Vios = ± 6 Vf = ± e V21. 

m = ± | • (4.582) = ± ± 3 . 927 . 

mi = + 3.927; m 2 = - 3.927. 


is read “m one.” The numeral 1 is called in such cases a sub¬ 
script. u m i ,> is read “m two.” These subscripts are used in this 
solution to distinguish between the two roots of the quadratic. 

Check. — In cases such as this, it is better to check by going over the 
solution a second time. Great care must be taken, however, for it is 
easy to overlook an error. 


EXERCISE 127 

1. 11 X 2 - 99 = 0 

2. 5x 2 - 125 = 3 x 2 -27 

3 . 13 c 2 + 8 = 88 - 7 c 2 

4 . 5 (m — 15) -f- to (3 to — 5) =0 
6. 7 (t - 2) + t(t - 5) = 2 t 

6. (to -f 3) (to — 1) = to(2 —3 m) 

7. (c — 2) 2 = (2 c - 8)(c + 2) 

8 . (3 to - 2)(3 to + 1 ) = 3(3 - to ) 

9 . (2x — 3) 2 =2(z -5)(z - 1) 

10. (5 c + 1 )(c - 3) - 2(2 - 7 c) = 0 



262 


ALGEBRA 


-L _ J_ =£ 

2 s 3 s 4 

4 r — 1 __ r + 2 
2r + 1 ~ Sr +2 

a: + 3 a- - 3 = „ 
j-3 2+3 d 

4 y 2 + 5 3 ? / 2 + 1 _ 5 y 2 + 15 

3 6 9 

—-f--— = 7 

r + 1 2r - 1 

a: 2 + x + 1 _ x 2 + 5 x + 7 
a; — 1 a: + 1 

17. Solve for x : a — 2 cx 2 = 3 b. 

Solution. — 1. a — 2 cx 2 = 3 b. 

2. So - 2 cx 2 = Sb - a. 

3. D. 2c a? =2_^3_6 

2c 

4. x = ± A/ a ~ 3 6~ __ , / 2 c(g — 3 &) _ ] 

^ 2 c > 4 c 2 J 

5 - ^ * = ± 2^ V2 c(a - 3 6). 

Solve for £: 

18. a 2 + x 2 = m 19. cx 2 = d 20. a + bx 2 = c 

21 . Solve s = \gt 2 for t. 22. Solve/ = — for a. 

r 

176. A right triangle is a triangle which has a 
right angle for one of its angles; as triangle ABC, 
in which angle B is a right angle. 

The side opposite the right angle is the hypote¬ 
nuse ; as, side AC. The side BC is the base and AB 
is the altitude . 



11 . 

12 . 

13. 

14. 

16. 

16. 






















QUADRATIC EQUATIONS 


263 


In a right triangle, the square of the hypotenuse equals the sum 
of the squares of the other two sides. 

Thus, b 2 = a 2 + c 2 . 

To verify this fact, draw a right triangle with BC 3 inches and AB 
4 inches; measure AC. Substitute the lengths of the sides in the 
equation b 2 = a 2 + c 2 . 

EXERCISE 128 

Carry out all computations in this set to two decimal places. 

1. Find the altitude of a right triangle whose base is 15 feet 
and whose hypotenuse is 30 feet. 

Solution. — 1. The formula is b 2 = a 2 + c 2 . 

2. In this problem a = 15, b = 30, and c = ? 

3. 30 2 = 15 2 + c 2 . 

(Complete the solution.) 

2 . Find the base of the right triangle whose hypotenuse is 
35 feet and whose altitude is 20 feet. 

3. If the diagonal of a rectangle is 65 inches and the base 
of it is 3 times the altitude, what are the dimensions of the 
rectangle ? 

4. a. The altitude of a rectangle is a inches. The base is 
3 times as long. What is the length of the diagonal ? 

b. Using the result of part a as a formula, how long is the 
diagonal when a = 5? a = 8 ? 

6 . Solve the formula b 2 = a 2 + c 2 : a. for b ; b. for c. 

c. Express the results of parts a and b in words. 

6 . Triangle ABC at the right is an isosceles 
triangle. AD is perpendicular to BC ; it is the 
altitude to the base BC. 

a. Compare BD and DC by measurement. 

b. Suppose AB = 15 inches and BC is 20 inches. Find AD. 

7 . If the equal sides of an isosceles triangle are each 25 inches 
long, and the altitude is 15 inches, find the base. 





264 


ALGEBRA 


8. a. If the equal sides of an isosceles triangle are each m 
inches long and the base is 2 n inches long, derive the formula 
for the altitude. 

b. Using your formula, find the altitude when the sides 
measure 20 inches and the base 14 inches. 

9 . An equilateral triangle has all three sides equal. What 
is the altitude of the equilateral triangle whose sides are 12 
inches long? 

10 . Find the altitude of the equilateral triangle whose sides 
are s units long. 

11 . Recall the formula for the area of a circle. (See page 
29.) What is the area when the radius is 8 inches? 

12. What is the radius of the circle which incloses an acre 
of ground (43,560 sq. ft.) ? 

13 . Solve the formula A = nr 2 for r, in simplest radical 
form: 

a. letting tt = 3y; b. without substituting a value for t. 

14. The volume of a circular cylinder is given by the 
formula V = ir r 2 h, where r — the radius, and h = the 
altitude. 

Find V when r = 7 and h = 20. 

15 . Find r when V = 1100 and h = 14. 

16 . Solve the formula of Example 14 for r. 

17. The formula for the area of the surface of a sphere is 

S = 47rr 2 . 

Find S when r = 15 in. 

18. Find in inches the radius of the sphere whose surface 
area is 1 square yard. 

19 . Solve the formula S = 4 tit 2 for r, without substituting 
its value for tt. 

20. Solve the formula V = 4 tt r 2 h for r. 






QUADRATIC EQUATIONS 


265 


177. A complete quadratic equation is an equation having 
only one unknown, in which the unknown does not appear in 
the denominator of the equation, and in which the first and 
second powers of the unknown do appear. 

Thus, 2 x 2 — Sx — 5 = 0 is a complete quadratic equation. 

178. Solution by factoring was taught on pages 148 and 149. 

EXERCISE 129 

Solve for the unknown number: 

1. y 2 — 6 y = 55 


3. 3 w 2 — w -f 1 = 0 


2. 8 a 2 —5a — 3=0 

4. A " 1 = 1 

Solve for x : 

6. 7 x 2 - 10 bx + 3 b 2 = 0 

„ 3 x 2 7 xt 

8 ' To "X 

6. | + 6c 2 =^f- 

x* _rx _ 

9 ' 10 6 

7. e* 2 — 47 ™* + m’ 

5 

10. £- 22 . 
5 15 


nr 1 


2 

llr 2 

15 


Solve for the unknown number: 


11. 

5 

— X 

9 

14. 

1 

-f * 

_ 5 

1 

— X 

5 — 4 x 

1-2 t 

+ 1 - 4 t 

6 

12. 


8 

- 12 -5 

15. 

r + 1 

2 r — 1 


y 

+ 1 

y- 2 

r — 1 

4r +1 


13. 


9 

3 _ 1 

16. 

2^-5 

5 +1 . 

= 0 

z 

+ 3 

22-3 2 

3s + 5 

55-1 


179. Solution of complete quadratics by completing the 
square. 

Preparation. — 1. Find : a. (x + 4) 2 ; 

c. (2 + i) 2 ; d. (w - i) 2 ; e. (r - f) 2 ; 


b. (y - 5) 2 ; 

/• (» - f ) 2 - 


















266 


ALGEBRA 


2. Make a perfect square trinomial of each of the following 
expressions (see page 249), and give its square root. 

a. x 2 -f 8 x ; b. y 2 — 14 y; c. w 2 — 20 w ; d. m 2 — § m. 

3. Notice that the coefficient of the second power in each of 
these expressions is 1; that the term added in each case is the 
square of one half the coefficient of the first power; thus, in 
a, you add (-f) 2 . 

This is called completing the square. It is used in solving 
quadratics. 

Example 1. — Solve x 2 — 12 x + 20 = 0 by completing the 
square. 

Solution. — 1 . S20 x 2 — 12 x = — 20 . 

2. We can complete the square of the left member by adding (^-) 2 
or 36. But then we must also add 36 to the right member. 


3. A 36 

x 2 - 12 x + 36 = - 20 + 36. 

4. 

/. (x - 6) 2 = 16. 


5. V 

x — 6 = =*= 4. 

Read Note, page 260. 

6 . 

.*. Xi — 6 = + 4, or xi = 10. 

This is one root. 

7. 

£2 — 6 = — 4, or X2 = 2. 

This is the other root. 


Note 1. — For meaning of x\ and x 2 , see Example, page 261. 

Check. — a. These roots check when substituted in equation 1. 

b. A second convenient method of checking. 

Add the roots: 10 + 2 = 12 . Observe that this is the negative of the 
coefficient of x in the original equation. 

Multiply the roots: 10 • 2 = 20. Observe that this is the term free 
from x in the original equation. 

This method of checking may be used only when the coefficient of the 
second power is 1 . 

Example 2 . — Solve x 2 — 6 x — 5 = 0 . 

Solution. — 1 . A 5 x 2 — 6 x = 5 . 


2 . 



(- 6) + 2 = - 3; 

(— 3) 2 = 

+ 9. 

3. 

a 9 


x 2 - 6 x + 9 

= 5 + 9. 


4. 



•*. (*-3) 2 

= 14. 


5. 



x - 3 

= ± Vl4. 


6 . 


When x 

- 3 = + vTi, xi = 

3 + Vl4. 

These are the roots 



When x 

1 

CO 

II 

1 

<, 

$ 

II 

= 3 - Vl4. 

in radical form. 


QUADRATIC EQUATIONS 


267 


7. x x = 3 + Vl4 = 3 + 3.741 = 6.741. These are the roots 

xt = 3 — Vl4 = 3 — 3.741 = — .741. in decimal form. 
Check. —Method a. When x = 6.741, 
does (6.741) 2 - 6(6.741) - 5 = 0? 

Does 45.441 - 40.446 - 5 = 0? 

Does 45.441 - 45.446 = 0? Almost. 

Since the roots are correct only as far as the third decimal place, 
we do not expect this root to check exactly. 


Method b. Adding 

the roots 6.741 



- .741 

This gives — (— 6 ) 


+ 6 . 

as it should. 

Multiply the roots: 

6.741 



- .741 

This gives — 4.99 + . 


6 741 

It should give — 5. 


269 64 

The result is near enough 


4 718 7 

to indicate that the 


- 4.995 081 

roots are correct. 


The only roots which will check absolutely are the radical roots. 

Rule. — To solve a quadratic equation by completing the 
square: 

1. Simplify the equation; transpose all terms containing the 
unknown number to the left member, and all other terms to 
the right member so that the equation takes the form 

ax 2 + bx = c. 

2. If the coefficient of x 2 is not 1, divide both members of the 
equation by it, so that the equation takes the form 

x 2 -f px = q. 

3. Find one half of the coefficient of x; square the result; 
add the square to both members of the equation obtained in 
Step 2. This makes the left member a perfect square. 

4. Write the left member as the square of a binomial; ex¬ 
press the right member in its simplest form. 

5. Take the square root of both members, writing the double 
sign, =±=, before the square root in the right member. 







268 


ALGEBRA 


6. Set the left square root equal to the + root in the right 
member of the equation in Step 5. Solve for the unknown. 
This gives one root. 

7. Repeat the process, using the — root in Step 5. This 
gives the second root of the equation. 

8. Express the roots first in simplest radical form, and then 
in simplest decimal form. 


EXERCISE 130 


Solve by completing the square. Check 1-6. Check others 
as directed by your teacher. 


1 . z 2 - 8 x + 15 = 0 

2. y 2 + 8 y + 12 = 0 

3. z 2 + 2 2 - 35 = 0 

4. z 2 - 2x - 1 = 0 


5. y 2 —4y + l= 0 

6. z 2 = 6 z — 4 

7. w 2 + 10 w + 22 = 0 

8. s 2 -J- 8 s -J- 10 = 0 


9. Solve the equation x 2 — 9 x + 7 = 0 . 
Solution. — 1 . S 7 x 2 — 9 x = — 7. 


2 . 

3. A^- 

4. 

5. 

6 . 

7. 

8 . Then 
and 


(-9) -i- 2 = -f; (-t)* = V* 
x 2 -9x+^ = ^-7. 

(x - f ) 2 = _ 

- + ^ 53 . 

2 

7 2 
_ 9 ± V53 


x - f 


xi = 


9 + V|3 _ 9 + 7.280 _ 16.28 


= 8.14 


*2 = 


9 - V53 _ 9 - 7.280 _ 1.72 


= . 86 . 


(Check this solution in one of the two ways taught.) 

10. X 2 + 3 x - 4 = 0 13. m 2 + m = 5 

11. y 2 + 5 y = 14 14. c 2 - 7 c + 3 = 0 

12. z 2 — 3 z + 1 = 0 16. r 2 - 9 r = 12 








QUADRATIC EQUATIONS 


269 


16. t 2 — 11 1 + 6 = 0 18. s 2 - 17 5 + 10 = 0 

17. a 2 - 15 a + 56 = 0 19. z 2 - 21 x - 22 = 0 


20. Solve Example 19 by the factoring method. 

21 . Solve the equation x 2 -f f x = f. 


Solution. — 1. 
2 . 

3. 

4. 


x 2 + | x = |. 

= (£) 2 — tt- 

z 2 +l*+Ar = -|+Ar- 
(*+i) 2 = if. 
(Complete the solution.) 


22. x 2 + -I x + -!• = 0 25. a: 2 + i x = 3 

23. x 2 - f a; - f = 0 26. a: 2 - f a: = 2 

24. a: 2 — § a: — 2 = 0 27. y 2 — § y — 3 = 0 

28. Solve the equation 3x 2 - 5 a: 4- 2 = 0. 

Solution. — 1. 3 x 2 — 5 x + 2 = 0. 

2. In order to make the coefficient of x 2 be 1, 

D 3 x 2 - f x + f = 0. 

(Complete the solution.) 


29. 2 x 2 + 5 x + 2 = 0 

30. 3 a: 2 - 10 x + 3 = 0 

31. 6 x* — x — 2 = 0 

32. 4 p 2 — 4 p = 1 

33. 5 r 2 = 2 r + 7 

34. 3 m 2 — 5 m — 4 = 0 


35. 7 x 2 —2a:—3=0 

36. 8 y 2 - 4 y - 1 = 0 

37. 9w 2 -6w)-4=0 

38. 6 t 2 - 8 t — 5 = 0 

39. 12 s 2 — 10 5 — 9 = 0 

40. 10 w 2 — 4 w — 3 = 0 


180. Solution of literal quadratics by completing the square. 

Example. — Solve for x the equation a: 2 + 3 ax + 4 b =0. 
+ 3 ax = —4 6. 

/3aV= 9a 2 

\2j 4 

n ^2 n ✓ 

3. 


Solution. — 1. 

2. \ of 3 a 


&y 

! 4-3ax4-^r = ^p-4 6. 

4 4 


270 


ALGEBRA 


5. 


4. 





_ ± V9 a 2 - 16 6 


= 9 a 2 - 16 6 


4 


2 


6 . 


x = 


_ - 3 a V 9 a 2 — 166 
2 ± 2 


7. 


= - 3 a ± — 16 6 

2 


o Tl _ - 3 a + V 9 a 2 - 16 6. r - 3 a - V 9 a 2 - 16 6 

8.x,- 2 --’ X2 -2-* 

Note 1. — This solution is now complete. 9 a 2 — 16 6 is not a 
perfect square and does not contain any square factors. 

Note 2. The solution can be checked by each of the methods pro¬ 
posed on page 266. Probably it is as satisfactory to check this solution 
by going over it carefully. 


EXERCISE 131 


Solve the following equations for x : 

1. x 2 + 2 ax + 3 = 0 8. x 2 — 2 mx = 1 — 2 m 

2. x 2 + 4 ax + 6 = 0 9. x 2 + 6 x = t 2 + 6 t 

3. z 2 - 6 bx - c = 0 10. x 2 - 4 ax = 6 2 - 4 ab 

4. 4a: 2 +8a: + w = 0 11. oz 2 + 4 x + 1 = 0 

6. 2 z 2 + a; - n = 0 12 . az 2 + 3 + 1 = 0 

6. 3 z 2 + 2 ax + 6 y 0 13. ckc 2 + 2 dx + m 0 

7. 3 re 2 + ax + 6 = 0 14. ax 1 + bx + c = 0 


181. Solution of quadratic equations by a formula. All 
quadratic equations having one unknown may be put in the 
form 


ax 2 + bx -f c = 0. 


For example, 2 z 2 - 3 x - 5 = 0 has this form. In it, 
a = 2ffb = — 3, and c = — 5. 

If we solve ax 2 + bx + c = 0 for x, the roots may be used 
as formulae for the roots of any quadratic equation. 














QUADRATIC EQUATIONS 


271 


Derivation of the formula: 


1 . 


ax 2 + bx + c = 

0 . 

2 . 

D a 

x 2 + - • x + — = 

0 . 



a a 


3. 

Sc 

x 2 + - • x = 

_ £ 


a 

a 

a 

4. 


lofH = A 

2 \a/ 2 a 

■<£ 

5 . 

A (£) 

* 2 +--X+-^ i = 

a 4 a 2 

6 2 

' 4 a 2 


6 2 

tn2 


£ 
a 

— \ac 


\ 2 a) 4 a 2 

x + — = ± ~ 4 ac 


2 a 
x 


2 a 


b Vfr 2 — 4 ac 
2 a 2 a 


X = 


b±vV — 4 ac 
2 a 


Read this solution carefully and be sure you understand it. 
But above all, memorize the formula. You can do so in two 
minutes. 

Example showing the use of this important formula. 

Solve the equation 2 x 2 — 3 x — 5 = 0 , by the formula. 
Solution. — 1. Comparing this equation with ax 2 + bx + c = 0 
a = 2, b = — 3, c = — 5. 

_ — b + Vb 2 — 4 ac 
2 a 


2 . The formula is x 


3. Substituting, x = 


- (- 3) ± V(- 3) 2 - 4(2) (- 5) 


2-2 


_ + 3 ± V 9 + 40 


x = + 3 ± ^49 or + 3 ± 7 

4 4 


3 + 7 10 5. , „ _ 3 - 7 _ - 4 _ , 

Xl = 4~ = ^ d ^ ~ ~4-4 


CTiecfc. — These roots check when substituted in the given equation. 




















272 


ALGEBRA 


EXERCISE 132 


1 . 

z 2 - 

7x + 6 = 

0 

11 . 

x 2 - 

- 2 x — 2 — 

0 

Hint. — a 

= i;& = - 

7; c = 6 . 

12 . 

x 2 4 

-2x - 1 = 

0 

2 . 

z 2 + 

2x - 15 = 

= 0 

13. 

2 a : 2 

+ 3 x — 1 ■ 

= 0 

3. 

2 a ; 2 

+ 5z+ 2 

= 0 

14. 

3 y 2 

1 

1 

to 

= 0 

4. 

3 y 2 

-7y + 2 

= 0 

15. 

2 z 2 

— 82 + 3 = 

= 0 

6 . 

2 z 2 ■ 

4- 7 z + 3 : 

= 0 

16. 

4 t 2 

- 9* + 2 = 

0 

6 . 

2w 2 

3 w — 2 

= 0 

17. 

5 c 2 

- 11 c -2 

= 0 

7. 

x 2 - 

2 a: - 35 = 

= 0 

18. 

6 m 2 

= 5 m + 3 


8. 

12 x 2 

+ 5x -2 

= 0 

19. 

7 d 2 

= 4 - 6o 


9. 

10 m 

2 — 11 m - 

-6=0 

20. 

8x 2 

+ 5 x - 1 = 

= 0 

10. 

24 x 2 

+ 14 a: - 

3=0 

21. 

5 x 2 

- 3a: - 1 = 

= 0 


182. Three methods of solving a quadratic equation have 
been taught: 

a. by factoring, in § 178, page 265. 

b. by completing the square, in § 179, page 266. 

c. by the formula, in § 181, page 271. 

Always try to use the factoring method first. 

If the expression is not easily factored, use one of the other 
two methods, preferably the third. 

Historical Note. —Greek mathematicians as early as Euclid were 
able to solve certain quadratics by a geometric method, about which 
the student may learn when he studies plane geometry. Heron of 
Alexandria, about 110 b.c., proposed a problem which leads to a quad¬ 
ratic. His solution is not given, but his result would indicate that he 
probably solved the equation by a rule which might be obtained from 
the quadratic by completing its square in a certain manner. Dio- 
phantus, 275 a.d., gave many problems which lead to quadratic equa¬ 
tions. The rules by which he solved his equations appear to have 
been derived by completing the square. He considered three separate 
kinds of quadratics. He gave only one root for a quadratic. 

The Hindu mathematicians, knowing about negative numbers, con¬ 
sidered one general quadratic. Cridharra gave a rule much like our 






QUADRATIC EQUATIONS 


273 


formula. The Hindus knew that a quadratic has two roots, but they 
usually rejected any negative roots. 

The Arabians went back to the practice of Diophantus in considering 
three or more kinds of quadratics. Mohammed Ben Musa, 820 a.d., 
had five kinds. He admitted two roots when both were positive. 
Alkarchi gave a purely algebraic solution of a quadratic by completing 
the square, and refers to this method as being a diophantic method. 

In Europe, mathematicians followed the practice of the Arabians, and 
by the time of Widmann, 1489, had twenty-four special forms of equa¬ 
tions. These were solved by rules which were learned and used in a 
mechanical manner. Stifel, 1486-1567, finally brought the study of 
quadratics back to the point that had been reached by the Hindus one 
thousand years before. He gave only three normal forms for the quad¬ 
ratic ; he allowed double roots when they were both positive. Stevin, 
1548-1620, went still farther. He gave only one normal form for the 
general quadratic, as do we; he solved this in both a geometric and an 
algebraic manner, giving the method of completing the square. He 
allowed negative roots. 

EXERCISE 133 


Solve the following equations by factoring if you can; other¬ 
wise by one of the other methods as your teacher directs. 


5. t - 


6 


6 . 


7. 


t - 3 

3 -,+* 

r 


= 0 
= - 2 


r + 1 


O _ _ = 1 

5-1 2 


1 

8 



5 


_ Q 



8. 

2 c 

4 _ ! 

1. 

X 


X 

+ 

3 

— O 




c + 5 < 

? — 3 

o 


3 




4 


4 

q 

3 m 

2 m _ j 

A. 

X 

+ 

4 


X 

- 1 




2 m + 1 

3m - 1 

Q 


2 




1 


_ 1 

10. 

2 

1 3 

o. 

X 

— 

1 


3 

x + 

1 

4 


4y + 1 

5 2 y - 1 

A 


2 




1 


2 

11. 

3z - 1 

2 z + 1 _ 3 

i k. 

y 

+ 

4 


y 

- 2 




x + 1 

x — 1 


12 . 


13. 


14. 


3 w 


w 


5 

7 - 3m 
2 

2c - 1 _ 
3c - 5 


w + 3 
3 + m 


= 1 
= 5 


2 — m 

3 c -}- 1 _1 

4c 6 




























ALGEBRA 


274 


15. 

5t + 1 

+-—| = i 

18. 

3r - 1 

2 - r 

= 2 


3 

t — 3 


5 + r 

2r + 1 


16. 

3 2 + 5 

2z + l_ 

19. 

5 c 

3 

o 

3b+2 

+ 2 2 

2c - 5 

c — 4 

Z 

17. 

2a — 3 

, « - 3 _ J 

20 . 

6 — 5 m 

3 + m 

3 

a + 3 

2a + 3 

2 

2 — m 

~ 2 


183. Problems solved by quadratics. Review § 49, page 
58, before solving the following problems. 


EXERCISE 134 

1 . If 3 be subtracted from 8 times a certain number, the 
result is 4 times the square of the number. What is the number ? 

2 . If 7 times the square of a certain number be increased 
by 3, the result equals 10 times the number. What is the 
number ? 

3. There are two consecutive integers whose product is 462. 
What are they ? 

4. The sum of the squares of certain two consecutive integers 
is 481. What are they? 

5. There are three consecutive even integers such that the 
square of the first, increased by the product of the other two, 
is 184. What are they ? 

6 . The sum of a certain number and its reciprocal is . 
What is the number ? 

Hint. — Read page 245. 

7. The length of a certain rectangle exceeds its width by 
8 feet. Its area is 308 square feet. What are its dimensions ? 

8 . The denominator of a certain fraction exceeds its numera¬ 
tor by 5. The sum of the fraction and its reciprocal is ££. 
What is the fraction ? 

9. The base of a certain triangle exceeds its altitude by 5 
inehes. Its area is 88 square inches. What are its dimensions ? 













QUADRATIC EQUATIONS 


275 


10 . The base of a certain rectangle is 9 feet longer than its 
altitude. Its area equals that of a square whose side is 20 feet. 
What are its dimensions ? 

11 . The area of a certain rectangle is 660 square feet. The 
sum of its base and altitude is 52 feet. What are its dimensions ? 

12. The total area of certain two squares is 369 square feet. 
The side of one exceeds the side of the other by 3 feet. How 
long is the side of each? 

13. The area of the main waiting room of the Union Railway 
Station at Washington, D. C., is 28,600 square feet. The sum 
of its length and width is 350 feet. What are its dimensions ? 

14. The altitude of a certain right triangle is 5 feet more than 
its base. The hypotenuse is 25 feet. What are its base and 
altitude ? 

15. The sum of the base and altitude of a certain right triangle 
is 17 feet. Its hypotenuse is 13 feet. What are its dimensions ? 

16. A lawn is 60 by 80 feet. How wide a strip must be cut 
around it when mowing the grass to have cut half of it? 

Hint. — Referring to the figure, it is clear that 
if w = the number of feet in the width of the 
border cut, then the dimensions of the uncut part 
of the lawn are (60 — 2 w) and (80 —2 w). 

Hence, (60 — 2 w) (80 — 2 w) = \ • 60 • 80. 

Complete the solution. 

17. A boy is plowing a field whose dimensions are 20 rods 
and 40 rods. How wide a border must be cut around it in order 
to have completed -f of his plowing ? 

18. If a certain number be increased by 3 times its reciprocal 
the sum is 12 \. What is the number ? 

19. The numerator of a certain fraction is 5 less than its de¬ 
nominator. If 3 be added to both numerator and denominator, 
the new fraction exceeds the original fraction by x ^. What is 
the fraction ? 


w Jw 

w 1 w 

1 

1 

80-2w | 

!s 

i 

i 


i 

i° 

i 

jvo 

l 

1 

w » 

1 

' !w" 

wj w 









276 


ALGEBRA 


20. A man traveled 84 miles by automobile at a certain rate. 
By increasing his average rate 7 miles per hour, he made the 
return trip in 1 hour less time than the time going. What was 
the rate going and returning ? 

21. Suppose a motor boat, traveling at the rate of 9 miles 
per hour in still water, goes 24 miles downstream, and back 
again in a total of 6 hours. WTat is the rate of the current of 
the stream? 

22. An auto trip to a town 100 miles away required hours 
for the round trip. Going, the party traveled 5 miles more per 
hour than they did returning. What was their rate going and 
returning ? 

23. The denominator of a certain fraction exceeds its numera¬ 
tor by 7. If the numerator be doubled, and the denominator be 
increased by 2, the sum of the resulting fraction and the original 
fraction is What is the fraction ? 

24. An automobile made a trip of 84 miles, of which 9 miles 
were inside village limits. The average rate outside villages 
was 12 miles per hour more than inside. If the trip required 
3 hours, what was the rate inside and outside villages ? 

25. A is now five times as old as B. A’s age 5 years from 
now, divided by B’s age then, forms a fraction which is f of 
the fraction found by dividing A’s age 15 years from now by 
B’s age at the same time. What are their ages now ? 

EXERCISE 135 

Miscellaneous Review 

Note. — The following examples were proposed as part of a recent 
examination given to pupils in Regents High Schools of New York. 

1. Divide a 4 + 2 a 3 — 6 a 2 + 26 a — 15 by a 2 + 4 a — 3. 

Check, letting a = 2. 

2. Factor: a. a 2 — 2 a — 48 c. 36 m 2 + 60 mn -f- 25 n 2 

b. 100 z 4 -49 y 6 d. 12 c 2 + 7 c - 12 


QUADRATIC EQUATIONS 


277 


3 . Simplify: (4--^). ( 8 —i£_^> 

4. Solve and check the system : { ® x V ~ 

\x + 2 y = .8 

5. Simplify : 2V108 - 6 aV5J + V 3(8 x - l) 2 . 

6. Solve and check : = x + x ^ • 

2 a; - 3 

E 

7. a. Solve for r : (7 = —-— 

R + r 

b. Find r to the nearest tenth when C = 15.6, R = 4.3, and 
E = 110.7. 

8. Find to the nearest tenth the roots of x 2 + 2 x = f . 

9. In a classroom there are 36 desks, some single and some 
double; if the seating capacity of the room is 42, how many 
desks of each kind are there? 

10 . In a certain high school, there were in the: 

First year, 690 pupils, Third year, 420 pupils, 

Second year, 750 pupils, Fourth year, 300 pupils. 

Represent this information by a bar graph. See page 208. 

11. The numerator and denominator of a certain fraction 
are in the ratio 2:3. If 3 be subtracted from the numerator, 
and 6 from the denominator, the value of the fraction becomes 
■J. What is the fraction ? (Hint. — Let 2 z be the numerator 
and 3 z be the denominator.) 

12 . a. The length of a room is L yards, the width is W yards 
and the height is H yards. ( 1 ) Represent the total surface of 
the four side walls ; ( 2 ) of the ceiling. 

b. From a board b feet in length, a piece c feet long was cut off 
from one end, and a piece d inches long from the other. How 
many inches in length was the piece which remained ? 








278 


ALGEBRA 


SUPPLEMENTARY TOPICS 


184. Examples and problems involving literal quadratics. 

Do these examples as you did the ones in Exercise 133, page 
273, using the formula method when you cannot use the factor¬ 
ing method of solution. 


EXERCISE 136 


Solve for x: 

1 . x 2 — 4 cx — 5 c 2 = 0 

2 . 3 x 2 — rx — 2 r 2 = 0 

3. x 2 + 2 xp + c 2 = 0 

4 . 3 x 2 + 2 mx — n = 0 

5. s = ax + | gx 2 
(From physics) 


6 . x 2 - Px + (P 1 ) = 0 

7. x 2 + (a + l)x + a = 0 

8 . x 2 — (a — fe)x — ab = 0 

9. cx 2 — (c + d)x + d = 0 

10. 7rx 2 + 7rxZ —5=0 

(From geometry) 


11 . a. The square of a certain number equals the sum of that 
number and n. What is the number? 

b. Using your results as a formula, find the number, when 
n = 12 . 

12. a. If the square of a certain number be increased by n 
times the number, the result is to. What is the number? 

b. Using your results as a formula, what is the number when 
n = 8 and to = 9 ? 


13. a. There are two consecutive even integers whose product 
is p. What are they ? 

b. Using your results as formulae, find the integers when 
p is 48. 

c. Suppose p = 9. Do the results have real meaning accord¬ 
ing to the statement of the problem ? 

14. a. The length of a certain rectangle exceeds its width by 
r feet. Its area is s square feet. What are its dimensions ? 

b. Using your results as formulae, find the dimensions when 
r = 12 and s = 13. 

c. Also when r = 10 and s = 100 . 


QUADRATIC EQUATIONS 


279 


15. a. The sum of a certain number and its reciprocal is M. 
What is the number ? 

b. Using your results as formulae, find the number when M 
is 

16. a. The sum of the base and altitude of a certain rectangle 
is p feet, and the area of it is A square feet. What are its 
dimensions ? 

b. Using your results as formulae, what are the dimensions 
when p — 40 and A = 375 ? 

185. Quadratic equations having two unknowns. 

Example 1 . —Find two integers (whole numbers) such that the 
sum of their squares is 90, and that the larger exceeds twice the 
smaller by 3. 

Solution. — 1. Let l = the larger integer 

and s = the smaller. 

2. /. Z 2 + s 2 = 90 (1) 

3. and l = 2 s + 3. (2) 

4. Eliminate l by the substitution method of elimination. (See page 
229.) 

From (2) l = 2 s + 3. 

Substituting in (1), (2 s + 3) 2 + s 2 = 90. 

5. /. 4 s 2 + 12 s + 9 + s 2 = 90, or 5 s 2 + 12 s - 81 = 0. 

6 . Factoring, (5 s + 27) (s — 3) = 0. 

7. Si = 3, and S 2 = — 

8 . When s = 3, Z = 2- 3+ 3 = 9. 

9 . — 2^. is unsatisfactory as a solution of the problem, since integers 
were desired. 

Check. — When s = 3 and l = 9, s 2 + Z 2 = 9 + 81 = 90. 

Also 9 = 2-3 + 3. 

When there are two unknown numbers, two independent 
equations must be given. (See page 226.) 

In the following list, one of these equations is a quadratic 
equation and the other is a linear equation. (See page 225.) 


280 


ALGEBRA 


Example 2 . — Solve the system 


* + V = 2 
xy = — 15 


( 1 ) 

( 2 ) 


Solution. — 1. Solve (1) for y in terms of x. y — 2 — x. 

2 . Substitute in (2), x{2 — x) = — 15. 

3. 2 x - x 2 = - 15. 

4. .*. x 2 -2x - 15 = 0, or (x + 3)(s - 5) = 0. 

5. xl — — 3, and x 2 = 5. 

6 . Since y = 2 — x,when z = — 3, y = 2 — (— 3) = 5; 

when rc = 5, y = 2 — 5=— 3. 

Check. — Each solution (see page 225) should be checked by sub¬ 
stituting it in both equations, (1) and (2). 


EXERCISE 137 


1 . 

2 . 

3. 

4. 

6 . 


f X - y = 3 
1 xy = 10 
fx+y = -2 
i zy = - 24 
f 2 a; + y = 5 
1 xy = 3 
{ a 2 + 6 2 = 13 
[a—26 = —4 

f x 2 ~ V 2 = - 7 
j 2 z - y = -10 


f erf « - 10 

* l2c + 3rf = 11 
138+t-O 

[ 9 s 2 + 4 t 2 = 5 
g f 4 m 2 -f n 2 = 10 
{ 2 m + n = 4 
f z 2 - 2 xy - y 2 = 17 
‘ \x -2y = 7 
f r 2 + + s 2 = 5^ 

* lr - 2s = 1 


11 . Find two numbers whose sum is 10 and whose product 
is 21 . 

12 . Find two numbers whose sum is 11 and the sum of whose 
squares is 61. 

13. One number exceeds another number by 3. The square 
of the larger exceeds twice the square of the smaller by 14. 
What are the numbers ? 

14. One number exceeds 3 times a smaller number by 1 . The 
larger, increased by the square of the smaller, is 11 . What 
are the numbers ? 


QUADRATIC EQUATIONS 


281 


16. The perimeter of a certain rectangle is 30 inches. Its 
area is 36 square inches. What are its dimensions ? 

16. The units’ digit of a certain number exceeds its tens’ 
digit by 5 . If the number itself be multiplied by its tens’ digit, 
the product is 114. What is the number? 

17. The sum of the squares of the two digits of a number is 
37 . If the digits be reversed, 4 times the new number exceeds 
the original number by 3. What is the number ? 

18. The perimeter of a certain parallelogram is 36 inches. 
The square of the longer side exceeds 6 times the square of the 
shorter side by 19 inches. How long are its sides? 

19. One of two angles exceeds the other by 10 °. If each of 
the angles be multiplied by its complement (see page 23), the 
two products are equal. What are the angles? 

20 . Two angles are supplementary. The square of the larger 
exceeds 100 times the smaller by 2000. What are the angles ? 

21. The rate of a certain passenger train exceeds that of a 
freight train by 15 miles per hour. The passenger train re¬ 
quires 3 hours less than the freight train for a trip of 140 miles. 
What are their rates ? 

22 . A and B together can do a certain piece of work in 8 days. 
It takes B 12 days longer than A to do the work alone. How 
long does it take each to do it alone ? 

23. The rate of a certain passenger train exceeds that of a 
freight train by 18 miles per hour. The passenger train re¬ 
quires 1 hour less time for a trip of 160 miles than the freight 
train requires for a trip of 110 miles. What is the rate of 
each? 

24. It takes A 2 days longer to do a certain piece of work 
than it does B. They can do the work if A works alone 5 
days and B works 4 days. How long does it take each to do 
it alone ? 


282 


ALGEBRA 


GRAPHICAL SOLUTION OF EQUATIONS WITH ONE VARIABLE 

186. Definition of function. Consider 3 x — 12. For every 
value of x, there is a definite value of 3 x — 12. 

3 x — 12 is said to be a function of x. 

One number (or number expression) is a function of another 
number when there is a definite value of the first for every value 
of the second. 

Thus, 5 • b represents the area of a rectangle of base b and altitude 5. 
For every value of b there is a definite value of the area of 5 b. Conse¬ 
quently, the area of a rectangle of altitude 5 is a function of the base. 

Every expression involving z is a function of x. 

Thus, 2 £ is a function of a:, — a first degree function, 
x 2 is a function of x, — a second degree function. 
x 2 — 3 x + 2 is a function of x, — a second degree function. 

187. Graph of a function. Consider again the function 
3 x — 12. As a matter of convenience, let y represent the 
value of 3 x — 12. 


When 

x = 

0 

2 

3 

4 

5 

6 

etc. 

Then y (or 3 x — 

12) = 

- 12 

- 6 

- 3 

0 

+ 3 

+ 6 

etc. 


Use each pair of 
numbers as the coor¬ 
dinates of a point; 
thus, the point (2, —6). 
The abscissa, 2, is the 
value of x; and the 
ordinate, — 6, is the 
corresponding value of 
the function. 

Connect the . points 
by a smooth line. The 
graph is at the left. 











































































































QUADRATIC EQUATIONS 


283 


188. Equations having one unknown can be solved graph¬ 
ically. 

Example 1. — Solve the equation 3 x = 12 graphically. 

Solution. — 1. 3 x = 12. 

2. S12 Sx -12 = 0. 

3. Draw the graph of the function 3 x — 12. 

(This was done in § 187, page 282. Look at that graph.) 

4. Observe point A. For it, the value of the function (that is, the 
value of y) is 0. The value of x which gives that value of y is 4. That 
is, when x = 4, 3 x — 12 is 0. 

5. 4 is the required root of the equation. 

Note 1. —Remember that we seek a value of x which makes 3 x — 12 
equal zero when we solve the equation 3 x — 12 = 0. That value of x 
in the graphical solution is the abscissa of the point where the graph 
crosses the x-axis. 

Note 2. — This graphical method of solving an equation having only 
one unknown is studied further in more advanced courses in algebra. 

Rule. — To solve an equation containing one unknown graph¬ 
ically : 

1. Clear the equation of fractions, collect like terms, and 
transpose all terms to the left side of the equation. 

2. Represent by y the function obtained in Step 1. 

3. Select values of the variable given in the equation and 
find the corresponding values of y. 

4. Use each pair of values obtained in Step 3 as the co¬ 
ordinates of a point; locate the points; draw the graph, mak¬ 
ing the vertical axis the y-axis, or the function-axis. 

5. Each point at which the graph crosses the horizontal axis 
represents a zero value of y, and the abscissa of such a point 
is the value of the variable which gives the zero value of y. 
Each such abscissa is a root of the given equation. 


284 


ALGEBRA 


Example 2. — Solve the equation — — - = - graphically. 

3 3 a: 


Solution. — 1. 


* _ 1 = 2 
3 3 x 


2. M 3 X x 2 — x = 6, or x 2 — x — 6 = 0. 

Let y = the function x 2 — x — 6. 


When x = 

0 

+ 1 

+ 2 

+ 4 

+ 5 

- 1 

- 2 

- 3 

- 4 

Then y = 

- 6 

- 6 

- 4 

+ 6 

+ 14 

- 4 

0 

+ 6 

+ 14 


5. The graph follows. 



6. The graph crosses the horizontal axis at A and B. 

At A, x = — 2. At B, x = + 3. These are the roots of the 
given equation. 

Check. — When x = — 2 does —- 2 — 1 = 2 ? 

3 3-2* 

Does-f-| - l? Yes. 

Similarly, + 3 checks. 

Note. — Do not say “draw the graph of the equation.” Say “solve 
the equation graphically.” To do so, you draw the graph of a certain 
function derived from the equation. 


















































































































QUADRATIC EQUATIONS 


285 


EXERCISE 138 

Solve the following equations graphically: 

1. 3 x = 2 4. x 2 = 10 - 3 x 

2. x 2 = 25 6. x 2 — x — 12 = 0 

3. x 2 + x = 12 6. x 2 - 4 x + 4 = 0 

189. Imaginary roots of a quadratic equation. 

Example. — Solve the equation x 2 — 2 x -f 5 =0. 

Solution. — 1. Use the formula method of solution. 

a — 1 ; b — — 2 ; c = 5. 

- (- 2) =*= V(- 2) 2 — 4 • 1 • 5 


2 . 


3. 


x = 


2 • 1 


V4 - 20 2 =*= V- 16 


What does V— 16 mean? 

— 4 is not the square root of — 16, since (— 4) 2 = + 16. 

-f 4 is not the square root of — 16, since (-j- 4) 2 = + 16. 

No number, studied so far, will produce a negative result when it is 

squared. V — 16 is a new kind of number, — an imaginary number. 

An imaginary number is the indicated square root of a nega¬ 
tive number, as V — 6 5 a/ — 3 » — i* 

The numbers studied previously, in contrast to the imaginary 
numbers, are called real numbers. 

190. Every imaginary number can be expressed as the prod¬ 
uct of a real number and V — 1 . 

\/ — 1 is indicated by i, and is called the imaginary unit. 
Thus, V- 16 = Vl6(— 1) = ± W- 1 = ± 4 i. 


vCTo 2 = vV(- i) = ± a v - 1 = 

V- 5 = V5(— 1) = =*= V- 1 • 5 = 

Note. —The symbol i was introduced by Euler, one of the greatest 
mathematicians of the eighteenth century. 

















286 ALGEBRA 

EXERCISE 139 

Express the following in terms of i : 

1. V- 4 

2. A/tTg 

3. V- 169 


4. V- 81 


6. V- 64 


6. 

V- 

- 25 X 2 

11. 

V- 

1 

¥ 

16. 

V- 

■ 27 

7. 

V- 

- 36 m 4 

12. 

V- 

1 

Tfi" 

17. 

V- 

48 

8. 

V- 

- 121 re 2 

13. 

V- 

4 

18. 

a/- 

32 

9. 

a/- 

- 400 x 2 

14. 

V^7 

19. 

V- 

9 

re 

10. 

V- 

- 196 x 4 

15. 

V- 

12 

20. 

V- 

75 


21. Simplify y /— 

Solution. — A / _ 27 _ / 9 • 3 • (— 1) 3 

\ 4 A 4 = ^ 9 


4 = = i v ^=-| i V3. 

Note. — a/3 -1 really gives V 3 • i, but the factor i is always 

written in front of any radical which multiplies it. 


22. V 


1 . 


25. V - 

26. V- 32 


24. V - 7 

4 

31. Simplify — 
Solution. — 1. 

2 . 

3. 


27. V- 20 


30. V 


/ 

2 7 


ss 

/ 

2 7 


re 


12 5 

TT 


2 7. 
T 


‘4 


3 - 9 - (— 1) 


= - =b 3 iV 3 
2 ~2~ 

_ - 5 ± 3 iV 3 


Note. — The numbers in Examples 1-30 ar/called pure imaginaries 
The numerator of the result in Example 31 is called a complex number. 

32. i 

33. f 

34. - l ± a/^ 


_ 5 
9 


35. 


V- 


JLrfc \/~Z. 9~ 

4 V T6 


36. - | ± V- || 


38. - * ± V 
39 


7 

3~6 


-9 =b A,/— 2 7 

10 v TOT 


¥5 


40. 4 ± a/^T 


8T 


37. f * v'TTjj: 















































QUADRATIC EQUATIONS 


287 


191. Solving quadratics which have imaginary roots. 

Example. — Solve the equation x 2 + x + 2 =0. 
Solution. — 1 . a — 1; b = 1) c = 2. 

_ 1 d= Vl - 8 _ - 1 =*= V^7 _ - 1 =±= Wl 

2. x - 2 _ 2 2 

0 - 1 + W7. - l - iV7 


We shall also solve this equation graphically. 
Solution. —• 1. Let y = x 2 + x + 2. 


When x = 

0 

+ 1 

+ 2 

4-3 

- 1 

- 2 

- 3 

- 4 

then y = 

+ 2 

+ 4 

4-8 

4- 14 

4-2 

4-4 

4-8 

+ 14 


3. The graph has the same 
shape as the graphs obtained 
when solving other quadratic 
equations; but the graph does 
not cross the horizontal axis at 
all. Hence, y or x 2 + * + 2 is 
never zero for any real value 
of x. 

This is characteristic of 
the graph of a quadratic 
which has imaginary roots. 



EXERCISE 140 

Solve the following equations either by completing the square 
or by the formula. Draw the graphs of the first three equations. 
1. x 2 + z + 3 = 0 


2. x 2 2 x + 2 = 0 

3. x 2 — 32+4=0 

4. 2x 2 — z + 1 = 0 

5. 3y 2 -2y +2 = 0 


6. 2 a 2 — 3 a + 2 —0 

7. 3t 2 —4*+2=0 

8. 4 r 2 — 5 r + 2 = 0 

9. 7 s 2 + 65+3=0 
10. 6 c 2 - 5 c + 2 = 0 














































































XV. SPECIAL PRODUCTS AND FACTORING 


ADVANCED TOPICS 

192. Product of two polynomials reducible to the type forms 
(a + 6) (a — b ) and (ax + b)(cx + d) studied in Chapter VIII. 

You have learned that (a + b) (a - b) = a 2 - b 2 , and that 
this relation is true for all values of a and b. 

Thus, if a = 5 x and b = 2 y, 

(5 x + 2 y)(5 x - 2 y) = (5 rr) 2 - (2 y ) 2 = 25 rc 2 - 4 y\ 

If a = 17 and b = 13, (17 + 13) (17 - 13) = 17 2 - 13 2 

= 289 - 169 = 120. 

If now, a = (x + y) and b = (z + w), then again, 

{(* + y) + (z + w)} {( x + y) — (z + w )} = (x + y ) 2 — (z -f w) 2 
= (re 2 + 2 xy + y 2 ) - (z 2 + 2 zw + a? 2 ) 

= x 2 + 2 xy + y 2 — z 2 — 2 zw> — ie 2 . 

In this example, (x -f y) takes the place of a, and ( z + w ) takes the 
place of b. 

By properly grouping terms, certain polynomials can be 
changed to type forms whose product can be found mentally. 
Example. — Find (r - 5 + t - u)(r + s + t + u). 

Solution. — Trial a. 1. Grouping r and s, and t and u, 

(r — s + t — u)(r + s + t + u) 

= {(r -«) + (*- ^)}{(r + «)+(< + u )}. 

(r — s) appears in the first set of braces, and (r + s) in the second. 
These are not the same number expression. Therefore this is an unsatis¬ 
factory grouping. 

Trial b. 1. Group r and t, and s and u. 

(r — s + t - u)(r + s + t + u) 

= {(r + <)-(«+ u)}{(r + t) + (s + «)}. 

Now, (r + t) appears in both sets of braces, and also (s + u). 

2. This is the type form (a — 6) (a + b) where a = (r + t) and 
b = (s + u). 

3. {(r + t) — (s + ^)}{(r + t) + ($ + u )} 

288 


SPECIAL PRODUCTS AND FACTORING 289 


4. = (r + t) 2 - (» + u) 2 

5. = (r 2 + 2 rt + £ 2 ) — (s 2 + 2 sw + m 2 ) 

6. = r 2 + 2 + * 2 — s 2 - 2 sit - u 2 . 


Check. — Let r = 1, s = 2, * = 1, ii = 2. 
Then 

(r-s-f-i — w) = l- 2 + l- 2= — 2 ] 

(r + s + t + n) = 1 +. 2 + 1+ 2= 6 


and (- 2) • (6) = - 12. 


Also, r 2 2 rt + t 2 — s 2 — 2su — u 2 = 1+2 + 1—4 — 8 — 4 — 


- 12 . 


EXERCISE 141 


Find by use of the special products: 

1. \(a -b) +6H(a-6) -6} 

2. S(c + d) -3e}{(c + d) + 3 e\ 

3. \8 — (a + b)\\8 + (a + b)\ 

4. {2 r — (»-<)! {2 r + (* -<)} 

6. \ (a - 3 d) + 4 c| {(a - 3 d) - 4 c\ j 

6. (* .+ y + z)(x + y - z) 

7. (r - s - t)(r - s + t) 

8. (m — p + n)(m + p + n) 

9. (2 a — b + c)(2 a + b — c) 

10. (x 2 + x - l)(z 2 - x + 1) 

11. (a 2 — ab + 6 2 )(a 2 + ab + 6 2 ) 

12. (r - 2 5 + 3 t)(r + 2 s + 3 <) 

13. (a + b + c — d)(a + 6 — c + d) 

14. (z + y - z — w)(x - y + z - w) 

15. {m - r + n + p)(m + r + n - p) 

16. Find (x 2 + 1 + 2 x){x 2 - 5 + 2 *) 

Solution. — 1. (x 2 + 1 + 2 zRx 2 — 5 + 2 z) 

2. = {(x 2 + 2 ®) + 1} {{x 2 +2x) - 5}. 

3 (z 2 + 2 x) is in both braces; but 1 and 5 are different numbers. 
This is a product like (y + 1 ){y - 5), which equals y 2 - 4 y - 5 where 
(x 2 + 2 x) takes the place of y. 

(Continued on page 290.) 





290 


ALGEBRA 


4. {(x 2 + 2 x) + 1} {(x* + 2 x) - 5} = (x 2 + 2 a;) 2 - 4(x 2 + 2 x) - 5 

5. = (x 4 + 4 x 3 + 4 x 2 ) - 4(x 2 + 2 x) - 5 

6. = x 4 + 4x 3 + 4x 2 — 4x 2 — 8x — 5 

7. = x 4 + 4 X s - 8 x - 5. 


Check. — Let x = 1. 


x 2 + 1+2x = 1 + 1 +2 = 4 \ ... „ 

x 2 - 5 + 2 x = 1 —5+2 = — 2 / and ^ 

Also, x 4 + 4x 3 — 8x — 5 = 1+4 — 8 — 5 = 



Find: 

17. f(a + 6) + 2|f(a + 6) + 3} 

18. }(* - 2 /) - 3|f(x - y) + 7| 

19. J6 — (m + w)} [8 + (m + n)\ 

20. J a + (b + c) | j a + (6 + c) } 

21. {(x - y) - z\\(x -y) - z\ 

22. (a + b — 4)(a + 6 + 6) 24. (7 + p + g)(8 — p — q) 

23. (a + y + 5)(x + y — 8) 25. (r — s — t)(r — s — t ) 

26. (x + y - a)(x + y - z) 

27. J2x - (r + *)|{3x + (r + s)\ 

28. J2 x - y + z\\2x - y - 5 a} 

29. 53 m — 2(a + 6) j J2 m + 3(a + 6) J 

30. 54 p — 5 r — 5 sj 53 p + 2 r + 2 sj 


193. Factoring generalizations of the type forms studied in 
Chapter VIII. 

Example 1. — Just as x 2 — y 1 = (x + y)(x — y) 

so (m — n) 2 — 25 a 2 = {(m — n) — 5 a}{(m — n) +5 a} 

= {m — n — 5 a}{m — n + 5 a}. 
Example 2. — Just as x 2 — 3 x — 88 = (x — ll)(x + 8) 
so (a - 2 6) 2 - 3(o - 2 6) - 88 = {(a - 2 6) - 11 }{(0 - 2 6) + 8} 

= {(0 - 26 - 11}{a -26+8}. 


EXERCISE 142 

Factor the following expressions : 

1. (a - b) 2 - c 2 3. (2 a - 6) 2 - 9 c 2 

2. x 2 - (s - i/) 2 4. 16 s 2 - (t + w) 2 


SPECIAL PRODUCTS AND FACTORING 291 


5. (to — n ) 2 — 8(to — n) + 15 

6. (r + s) 2 + 4(r + 5 ) — 77 

7 . (2 f — w) 2 — 3 5(2 t — w) — 10 s 2 

8. 4 z 2 - 4 2 ( 2 / + z) + ( 2 / + z) 2 

9. 6 x 2 + x(y + z) - 2(y + z) 2 

10. 4(to + n) 2 — 5(to + n)p — 6 p 2 

11. 4(a + b) 2 — 12(a + b) + 9 

12. 16(c - d) 2 - 9(x - y) 2 

13. (a: 2 + y 2 ) 2 — 4 x 2 y 2 

14. (a 2 - 2 a) 2 + 4(a 2 - 2 a) + 3 

15. (9 a 2 + l) 2 - 36 a 2 

16. (x 2 + 2x) 2 + 2(x 2 + 2 a;) + 1 

17. (r 2 - 3) 2 - (r + l) 2 

18. ( x 2 + l) 2 — 4 x 2 

19. (x 2 +2x) 2 - 2{x 2 + 2 *) - 3 

20 . (a 2 - 3 a) 2 - 2(a 2 - 3 a) - 8 

194. Polynomials reducible to the difference of two squares. 

Example. — Factor a 2 — c 2 + b 2 — d 2 — 2 cd — 2 ab. 

Solution. — 1. It is natural to group -2 ab with a 2 , and that at once 
suggests grouping b 2 with them. 

... a 2 - c 2 + b 2 -d 2 -2cd-2ab 

2 . = (a 2 - 2 ab + b 2 ) - (c 2 + 2 cd + d 2 ) 

3 . = (a - b) 2 - (c + d) 2 

4 . = {(a - b) + (c + d)}{(a - b) - (c + d)} 

5 . = { a - b + c + d}{a - b - c - d}. 

Check. — Let a = 1, b = 2, c = 1, d = 2. 

0 2_ c 2 + b 2_ ( p_2cd-2a5=l-l+4-4-4-4 = -8. 

{a — 6 + c + d{{a — 5 — c — d} = {l —2 + 1 +2}{1 —2 — 1 — 2} 

= 2 (— 4), or - 8 . 

Note. — A polynomial of four terms can be factored in this way if 
three terms make a perfect square trinomial and if the fourth term is 
a perfect square monomial, preceded by a minus sign. 



292 


ALGEBRA 


EXERCISE 143 


Factor: 

1. a 2 + 2 ab + b 2 — c 2 

2. m 2 — 2 ran + n 2 — 4 

3. it 2 + 6 * + 9 - y 2 

4. t 2 — r 2 — 2 rs — s 2 

6. 4 c 2 — a 2 + 2 ab — b 2 
6. 9 x 2 — 6 xy — z 2 + y 2 


7. 16 a 2 + b 2 + 8 ab — c 


15. x 2 + 2/ 2 


8 . 9 r 2 — 4 s — 4 — s 2 

9. 25 t 2 - 4 z 2 + 10 t + 1 

10. 8 x + 49 y 2 — x 2 — 16 

11 . 6f + 16 ra 2 — 9 — t 2 

12. 36 -12 ab - a 2 - 36 5 2 

13. 81 z 2 + 20 w - 4 w 2 - 25 

14. 4 Z?/ - z 2 + it 2 + 4 y 2 
2 xy + 2 cd — c 2 — d? 


16. 6 ab + a 2 — x 2 + 9 5 2 — 4 a; — 4 

17. 4 a 2 + 1 — 6 2 — 4 a — 9 — 6 b 

18. r 2 — 9 5 2 + 25 + 6 st — 10 r — < 2 

19. 4: a 2 — x? + 2 xy — 12 ab — y 2 + 9 b 2 


20 . 2 cd + 25 a 2 


- 10 ab - d 2 + 6 2 


195. Trinomials reducible to the difference of two squares. 

Type form : x 4 + x 2 y 2 + y 4 
Example. — Factor 64 a 4 — 64 a 2 m 2 + 25 ra 4 . 

Solution. — 1. A perfect square containing 64 a 4 and 25 m A is 64 a 4 — 
80 o 2 ra 2 + 25 ra 4 . It can be obtained from 64 a 4 — 64 ahn 2 + 25 ra 4 by 
subtracting 16 a 2 ra 2 . Then we must also add 16 a 2 ra 2 to avoid chang¬ 
ing the value of 64 o 4 — 64 a 2 ra 2 + 25 ra 4 . That is 

2. 64 a 4 — 64 a 2 ra 2 + 25 ra 4 = (64 a 4 — 80 a 2 ra 2 + 25 ra 4 ) + 16 a 2 ra 2 

= (8 a 2 — 5 ra 2 ) 2 + (4 ora) 2 

But this is not factorable, since it is the sum of two squares. 

3. Another perfect square containing 64 o 4 and 25 ra 4 is 64 o 4 + 80 a 2 ra 2 
+ 25 ra 4 . It can be obtained from 64 o 4 — 64 o 2 ra 2 + 25 ra 4 by adding 
to it 144 a 2 ra 2 . Then we must also subtract 144 a 2 ra 2 to avoid changing 
the value of the given expression. That is 

64 o 4 — 64 o 2 ra 2 + 25 ra 4 
= (64 o 4 + 80 o 2 ra 2 + 25 ra 4 ) - 144 o 2 ra 2 

4. = (8 o 2 + 5 ra 2 ) 2 - (12 ora) 2 

5. = {(8 a 2 * + 5 ra 2 ) + 12 am) {(8 a 2 + 5 ra 2 ) — 12 ara} 

6. = (8 a 2 + 5 ra 2 + 12 ora} {8 o 2 + 5 ra 2 — 12 am} 


SPECIAL PRODUCTS AND FACTORING 293 


Check. — Let a = 1 and 6=1; and m = 1. 

64 a 4 - 64 a 2 m 2 + 25 m 4 = 64 - 64 + 25, or 25 
{8 a 2 + 5 m 2 + 12 am} {8 a 2 + 5 m 2 — 12 am} 

= {8 + 5 + 12} {8 + 5 - 12} 

= 25 • (1), or 25. 

Note. — If you are far sighted or fortunate you may try Steps 3 
and 4, first, and then Steps 1 and 2 are, of course, unnecessary, and do 
not appear in your solution. 


Factor: 


EXERCISE 144 


1. a 4 + a 2 b 2 + b 4 

2. m 4 + m 2 + 1 

3. x 4 + 4 x 2 + 16 

4. y 4 + 9 y 2 + 81 

6. z 4 + 6 zV + 25 w‘ 

6. r 4 + 3 rV + 36 .s 4 

7. 9 to 4 + 2 7« 2 f 2 + t 4 

8. 16 a 4 - 12 a 2 + 1 

9. 9 x 4 - 10 x 2 + 1 
10 . 9 a 4 + 2 a 2 b 2 + 6 4 


11 . 36 f - 16 1 2 + 1 

12. 25 x 4 - 19 xV + y* 

13. 4 a 4 + 8 a 2 + 9 

14. 9 x 4 + 20 x 2 + 16 

15. 4 a 4 + 11 a?b 2 + 25 b 4 

16. 9 c 4 - 34 c 2 cP + 25 d 4 

17. 16 a 4 - 28 aV + 9 b 4 

18. 9 r 4 - 43 r 2 l 2 + 49 t 4 

19. 16 x 4 + 4 xV + 25 t/ 4 

20. 4 m 4 + 19 m 2 » 2 + 49 n 


Factor similarly the following binomials : 

21. x 4 + 4 23. c 4 + 4 <f 4 25. r 4 + 64 t 4 

22. 4 a 4 + 1 24. 4 x 4 + y 4 26. x 4 + 324 


196. Polynomials factored by grouping. 

Example 1. — Just as ax + bx = x(a + 6) 

so a(m + n) + b(m + n) = (m + n)(a + 6). 
In this example (m 4- n) takes the place of x. 

Example 2. — Factor 6X 3 — 15x 2 — 8x + 20. 

Solution . — 1. 6x* — 15 x 2 — 8x4-20 

2. = 3 34(2 x - 5) - 4(2 x - 5) 

3. = (3X 2 - 4)(2x - 5). 





294 


ALGEBRA 


Check. — Let x = 1. 

6 x 3 — 15 z 2 — 8 x + 20 = 6 — 15 - 8 + 20, or 3. 

(3z 2 - 4)(2 x - 5) = (3 - 4)(2 - 5) = (- 1)(- 3), or + 3. 


EXERCISE 145 


Factor: 

1. a(x +y) - b(x + y) 

2. 2 m(r — s) 3 n(r — s) 

3. p{ 2 a — b) — q(2 a — b) 

4. 2 x(y — z) — 3 w(y — z) 

5. 5(m + x) — r(m + x) 

6. ac -f- ad + be -f- bd 


9. z 3 — 2 x 2 + x — 2 

10. 2 a 3 - 3 a 2 - 3 + 2 a 

11 . 2 a — b—2ac-\-bc 

12. 3 - 6 b 2 + 5 6 - 10 6 3 

13. 8 * 3 + 2 < — 4 * 2 — 1 

14. 15 am — 6 bm + 5 ax — 2 bx 


7. rt + rw — st — sw 15. a 3 — a 2 b + ab 2 — 6 3 

8. mx — my — nx ny 16. c 3 + c 2 d — d? — cd 

17. 24 a 3 - 20 a 2 - 5 + 6 a 

18. 3 mx + 6 nx + my + 2 ny 

19. 2 or 3 — 6 £ 2 / 2 + x 2 y —3 y 3 

20. 40 £ 3 — 42 — 35 z 2 + 48 x 

21. ax — ay az bx — by bz 

22. a# — bx — ay — az + by -f- bz 

23. 2 ac — 6c+4a — 26 + 6+ 3c 

24. 2 x* — x 2 y + 2 x 2 — xy -\- 2 x — y 

25. 2 am — 3 bm — 2 an + 3 bn — 6 bp + 4 ap 

26. 2 rx + 2 rz — sx + 3 sy — 6 ry — sz 

27. 10 ap - 8 aq + 15 cp + 4 - 12 cq - 5 


197. The sum or difference of like powers of two numbers. 


Development. — 1 . Make a table like the following and fill 
in the quotients and remainders. 


Note. — A good class exercise is to place this table on the black¬ 
board. Time can be saved by distributing the division problems 
among the members of the class. 


SPECIAL PRODUCTS AND FACTORING 


295 




Dividend 

Divisor 

Quotient 

Remainder 

a. 

Sum of odd 
powers 

X 3 + y 3 
x 5 + y 5 

& + y 3 

x h + y 5 

X + y 

X + y 

X - y 
x - y 



6 . 

Sum of even 
powers 

x 1 + y 2 
x 4 + y 4 
x 2 + y 2 
x 4 + y 4 

x + y 
x + y 
x - y 
x - y 



c. 

Difference of 
odd powers 

n. « 

1 1 1 1 
% % % % 

x - y 
x - y 
x + y 
x + y 



d. 

Difference of 
even powers 

*1. 

I l l 1 

x — y 

x — y 
x + y 
x + y 




Examination of the results will disclose the following rules. 

Rules. — I. When n is a positive integer. 

1. When n is odd, x n -f y n is exactly divisible by x + y, but 
not by x — y. 

2. When n is even, x n + y n is not exactly divisible by either 
x + y or x - y. 

3. When n is odd, x n — y n is exactly divisible by x — y, but 
not by x + y- 

4. When n is even, x n — y n is exactly divisible by x — y, and 
also by x + y. 

II. As to the quotients in all cases: 

1 . When x — y is the divisor, the signs of the quotient are 
all plus. 





















296 


ALGEBRA 


2. When x -f y is the divisor, the signs of the quotient are 
alternately plus and minus, the first term being plus. 

3. The exponent of x in the first term is 1 less than its ex¬ 
ponent in the dividend, and decreases by 1 in each succeeding 
term until it becomes 1 in the next to the last term, x does 
not appear in the last term. 

4. y does not appear in the first term. The exponent of y 
in the second term is 1, and increases by 1 in each succeeding 
term until it becomes 1 less than the exponent of y in the divi¬ 
dend. 

III. Two particularly important formulae. 

1. x 3 - t/ 3 = (x - y)(x 2 4 - xy + y 2 ) Rules 1-3 ; II-l, 3, 4. 

2. x 3 + y z = (x + y)(x 2 - xy ■f y 2 ) Rules 1-1; II-2, 3, 4. 

Example 1. — Factor 8 a 3 — 27 b 6 . 

Solution. — 1. 8 a 3 - 27 6 6 = (2 a) 3 - (3 6 2 ) 3 . 

2. This is the difference of two cubes, and may be solved by the 
formula III-l. 

.% 8 a 3 — 27 b 6 = (2 a - 3 6 2 ) [(2 a) 2 + (2 a) • (3 6 2 ) + (3 6 2 ) 2 1 

3. = (2 a - 3 6 2 )(4 a 2 + 6 ab 2 + 9 6 4 ). 

Check. — Let a = 1; b = 1. 8 a 3 - 27 6 6 = 8 - 27, or - 19. 

(2 a - 3 6 2 )(4 a 2 + 6 ab 2 + 9 6 4 ) = (2 - 3)(4 + 6 + 9), or - 1 • 19 

= -19. 

Example 2. — Factor 32 x 5 + 243 y 10 . 

Solution. — 1. 32 x h + 243 y w = (2 a;) 5 -f (3 y 2 ) 5 . 

2. This is the sum of two odd powers. By 1-1, and II-2, 3, 4, 

32 x 6 + 243 y 10 

= (2 x + 3 s/ 2 ) [(2 x) 4 - (2 x) 3 (3 2 / 2 ) + (2 x) 2 (3 i/ 2 ) 2 - 
(2x)(3j/ ! ) s + (3y 2 ) 4 ] 

3. = (2x + 3y 2 )(16x 4 — 8x 3 *3y 2 + 4a^*9j/ 4 —2x*27y 6 

+ 81 ?/ 8 ) 

4. = (2 x + 3 i/ 2 )( 16 x 4 — 24 x 3 ?/ 2 + 36 x 2 y A — 54 xy 6 + 81 y s ). 

Check. — Let x = 1; y = 1. Then 32 x 5 + 243 y 10 = 32 + 243, 

or 275. 

(2 x + 3 ?/ 2 )(16 x 4 - 24 x=y + 36 xh/ ~ 54 xy 6 + 81 y 8 ) 

= (2 + 3)(16 - 24 + 36 - 54 4- 81) = 5(55), or 275. 


SPECIAL PRODUCTS AND FACTORING 


297 


The method of factoring binomials of the form x n ± y n given 
in this section must be used frequently. However, if the prime 
factors of a binomial of this form are wanted, proceed as follows : 

Whenever the binomial is the difference of two even powers, 
treat it as the difference of two squares. 

Example 3. — x 6 - y* = (x 3 - y 3 )(x 3 + y 3 ) 

= (x — y){x 2 + xy + y 2 ){x + y)(x 2 - xy + 2/ 2 )- 

When the binomial is the sum of two powers, try to express 
it as the sum of two cubes; if not cubes, the sum of two fifth 
powers; etc. 

Example 4. — 1. x 9 + y 9 = (x 3 ) 3 + (y 3 ) 3 

2. = (x 3 + y 3 ) [(z 3 ) 2 - OrW) + G/ 3 ) 2 ] 

3. = (x 3 + y 3 )(x 6 - xy + y 6 ) 

4 . = (x + y)(x 2 - xy + y 2 )(x 6 - xPy 3 + y e ). 

Note 1. — Observe x 9 is thought of as Or 3 ) 3 ; similarly, think of x 10 

as (x 6 ) 2 or as (x 2 ) 5 ; x 6 as (x 3 ) 2 or (x 2 ) 3 ; x 12 as (x 4 ) 3 or as (x 3 ) 4 . 

Note 2. — In Example 3, prime factors of x 6 — y 6 are found. 

a. By Rule 1-4, (x - y) is a factor of x 6 - y 6 . Find the other 
factor by II—1, 3, 4. Is this factor prime? 

b. Referring to Example 3, above, find the product of (x + y)(x 2 - 
xy _j_ t/ 2 ) (x 2 + xy + y 2 ). Compare it with the second factor in part a. 

c. By Rule 1-4, (x + y) is a factor of x 6 - y 6 . Find the other 
factor by II-2, 3, 4. 

EXERCISE 146 

Factor completely the following expressions, if possible. 


1. 3? — 81/ 3 

11. X 4 + 16 y 4 

21. 243 + o4° 

2. b 6 - 27 c 3 

12. x 4 - y 4 

22. 32 a 5 + 1 

3. c 6 + 8 <P 

13. x 6 + y e 

o 

1 

1ft 

H 

CO 

<N 

4. 8 r 6 + 1 

14. x 8 + y 8 

24. r 3 ^ - 27 f 3 

5.8a 8 - 27 b 3 

15. x 8 - y s 

25. a 7 - b 7 

6. 27 to 3 + n 6 

16. x 10 + y 10 

26. m 10 - 243 n 5 

7. - 125 f 

17. x 10 - y 10 

27. -g- x 3 — y 6 

8 . 64 + m 9 

18. x 12 - y 12 

28. a 3 + * 6 3 

9. 27 o 3 + 64 b 3 

19. x 4 - 16 

29. ^ x 3 - if 

10. 8 c 6 - 125 <P 

20. x 5 - 32 

30. a 5 — 6 5 




298 


ALGEBRA 


Reduce to lowest terms : 


31 * 4 ~ yi 

33 ^ + y 3 

36 8c3 

3 ? - y* 

. ' X 5 + y b 

36 ' 16 c 4 — (f 4 

32. f+ 8 

34. ~ 16 

36. 1 +27 *; 

x 4 - 16 

to 6 - 32 

1 - 9 x 4 

Simplify: 



a 

OO 

H. 

1 

to 

4 x 2 + 12 x + 9 

to 

1 

CO 

4x2-9 

4 x 2 + 6 x + 9 

6 x 2 -j- 5 x — 6 

s« m3 + n 3 . 

m — n (m — nf 

+ mn 


OO. -- —y-— 

m 3 — n 3 m -f n (m -f nf — mn 
39 3 m 4 — 3 ?i 4 < 2 m 2 — 2 n 2 ^ cm 2 + 2 cmn 4~ c^ 2 
4 m 3 - 4 w 3 6 m 2 + 6 n 2 4 m 2 +4 mn +4n 2 


Factor: 

40. (x — yf + z 3 

41. (r - sf - 8 J 3 

42. z 3 + (a; — l) 3 

43. a 3 — 8(a — 6) 3 

44. 27 a 3 + 8(a - bf 


45. ( x - yf - (x + yf 

46. (a + 2 bf + (a - 2 6) 3 

47. (3 x - yf - (3 x + yf 

48. (m — 2 nf — (m + 2 nf 

49. ( x +2yf + (2 x+ yf 


SUPPLEMENTARY TOPICS 


198. The Remainder theorem makes it possible to find the 
remainder in certain division problems by a short process. 


It is known that dividend = divisor 
Suppose that a? + x 2 — 2 is divided 
z* + z 2 - 2 = (x - 2) • Q + R. 

Let x — 2 ; then: 

8 + 4- 2 = 0- Q + R. 

10 = 0 • Q + R. 

10 = R. 

That is, the remainder is 10. 

Check. — See solution on right. 


X quotient -f- remainder. 
by x — 2 ; then : 
rr 2 + 3 x + 6 
x z + t? — 2 1 x — 2 
x? — 2 x 2 
3 x 2 

3 x 2 — 6 x 

6 x — 2 
6x — 12 


10 = R. 






















SPECIAL PRODUCTS AND FACTORING 


299 


At the left, the correct remainder was obtained by substitut¬ 
ing 2 for x in the given expression. This suggests the 

Remainder theorem. — If a rational and integral polynomial 
involving x be divided by x — a, the remainder may be found 
by substituting a for x in the given polynomial. 

Note. — A polynomial is integral if it does not have any negative 
powers of the variable x, or any fractions with x in the denominator. 

A polynomial is rational if it does not have any fractional powers of 
x, or any x’s under radical signs. 

Proof. — 1. The polynomial (containing x) = (x — a) • Q + R. 

2. .*. The polynomial (x replaced by a) = (a — a) • Q + R- 

3. The polynomial (x replaced by a) = 0 • Q + R } or R. 

Example 1.—Find the remainder when x 4 — 3 x + 5 is 
divided by x — 3. 

Solution. — 1. Comparing x — a and x — 3, a must be 3. 

2. Substituting 3 for x in x A — 3x + 5, 

R = 3 4 — 3 • 3 + 5. .'.R = 81 - 9 + 5, or 77. 

Example 2. —Find the remainder when the divisor is x + 3. 

Solution. — 1. Comparing x — a and x + 3, a must be — 3. 

2. Substituting — 3 for x in x 4 — 3 x + 5, 

R = (_ 3) 4 - 3 • (- 3) + 5. .*. R = 81 + 9 + 5, or 95. 

EXERCISE 147 

Find the remainder when: 

1. x 4 — 2 x? + 5 is divided: a. by x — 1; b. by x -f 2. 

2. 3Z 3 — 2x 2 - 4z + l is divided: c. by a: -2; b. by 

x + 3. 

3. y 4 — 3 y 1 — 7 is divided: a. by y + 2; b. by y — 1. 

4. z 5 — 4 z 3 -f z — 30 is divided: a. by z — 2; b. by z + 1. 

5. 3 m 4 — 5 m 3 — m — 6 is divided: a. by m — 2; b. by 
m — 3. 

199. Synthetic division is a short process for finding the 
quotient as well as the remainder when a polynomial containing 
x is divided by a binomial of the form x — a. 


300 


ALGEBRA 


Consider the two solutions : 
Solution a 
5 x 2 + 4 x +6 
x-3\ 5x?-llx*-6x-10 
5 x 3 - 15 x 2 

4 x 2 — 6 x 
4 x 2 - 12 x 

6x - 10 
6x - 18 
+ 8 


Solution b 

x + 3\ 5tf-llx 2 - 6 x — 10 
5 +15 +12+18 

5 + 4 + 6 || + 8 
Quotient: 5 x 2 + 4 x + 6. 

Remainder: + 8. 


The method of performing the solution b : 

1. — 3 of the original divisor is changed to + 3. 

2. 5 x 3 4- x = 5 x 2 . Place 5 in the third line. 

3. + 3 • + 5 = + 15. Add the product, +15, to — 11. Place 

the sum, + 4, in the third line. 

4. + 3 • + 4 = + 12. Add the product, + 12, to — 6. Place the 
sum, + 6, in the third line. 

5. + 3 • + 6 = + 18. Add the product, + 18, to — 10. Place 

the sum, + 8, in the third line. 

6. The numbers 5, + 4, and + 6 are the coefficients of the quotient. 
Since 5 x 3 -r x = 5 x 2 , the full quotient is 5 x 2 + 4 x + 6. The last 
number of the third line, + 8, is the remainder. 

A partial explanation follows : 

1. In Step 1, — 3 is changed to + 3. This permits addition in Steps 
3, 4, and 5 instead of the customary subtraction. Thus, in solution a, 
when — 15 x 2 is subtracted from — 11 x 2 , the result is 4 x 2 ; in solution 
b, when + 15 x 2 is added to — 11 x 2 , the result is again 4 x 2 . 

2. In Step 3, + 4 below the line represents, first, the coefficient of 
the first term of the remainder as in solution a. When 4 x 2 is divided 
by x the quotient is + 4 x, so that 4 may properly be considered also 
the coefficient of the second term of the quotient. 


Example. — Divide 7 x 4 — 29 fix 2 — 3 f 4 by x + 2 t. 


Solution. — Change x + 2 £ to x — 2 t. 


x — 2 t 


7 x 4 + 0 *x 3 - 29 t 2 x 2 + 0 * 3 x - 3 t 4 

7 - 14 t + 28 e +2 t* - 4 t 4 

7 -14 t - t 2 + 2 t 3 1| - 7 t 4 


Quotient: 7 x 3 - 14 te 2 - fx + 2 J 3 . Remainder: - 7 t 4 . 















SPECIAL PRODUCTS AND FACTORING SOI 


Note 1. — When powers of x are missing, supply them with 
coefficients zero, as in this example. 

Note 2. — In §§ 198 and 199, two short processes for finding the 
remainder in certain division problems are given. Each is important. 
The second has the further advantage of determining the quotient as 
well. 


EXERCISE 148 

Divide by synthetic division: 

1. i 3 + 2x 2 -2x + 5byx-l 

2. y 3 - 4 y 2 + y + 6 by y + 1 

3. 2 x 3 — 3 x 2 + x — 8 by x — 2 

4. 3 2 J 4 + 4 z 3 — 12 2 2 -{- 8 z + 6 by z + 3 

6. x 4 + x 2 - 6 by x + 2 

6 . 3 t 4 - 8 Z 3 — 30 by t — 3 

7 . m 4 — 256 by m + 4 

8. 5 r 3 + 6 r 2 - r - 12 by r - 2 

9. x 4 + 2 mx 3 — 5 m 2 'x 2 + 12 m 4 by x — 2 m 
10. 2 i/ 5 - 15 oV + 11 a 3 y 2 - 21 a 5 by y + 3 a 

200. The factor theorem makes it possible to factor certain 
polynomials which are not factorable otherwise. 

a. Consider (x — 2)(x + 3) = x 2 — x — 6. 

Observe that the numbers 2 and 3, of the binomials, are 
factors of the last term 6, of the trinomial. If the polynomial 
to be factored is arranged in descending powers of one letter, 
then the last term always is the product of the last terms of 
the factors of the polynomial. 

Thus, — 12, of x 3 + 3 x 2 — 4 x — 12 is the product of the last terms 
of the factors of x 3 + 3 x 2 — 4 x — 12. 

Therefore, the last terms of the factors must be selected from 1, 2, 
3, 4, 6, and 12. For example, 5 cannot be the last term of any factor 
of x 3 + 3 x 2 - 4 x — 12. 


302 


ALGEBRA 


b. Factor rc 3 + x 2 - 2 by trial. 

Solution. — 1. The factors must end in either 1 or 2. 

2. Is x 1 a factor? If it is, the remainder must be 0 when 
re 3 + £ 2 — 2 is divided by x — 1. 

By the remainder theorem, 12=1 + 1- 2 = 0. (See page 299.) 

3- {x — 1) is a factor of x 3 + x 2 — 2. 

Factor theorem. — If a rational and integral polynomial 
involving x becomes zero when x is replaced by a, then the 
polynomial has x - a as a factor. 

Proof. — 1. Considering the polynomial the dividend and (x — a) 
the divisor, the remainder is the value obtained when a is substituted 
for x in the polynomial. 

2. By the theorem, this value is zero, and .*. the remainder = 0. 

3- x — a is a factor. 

When applying this theorem : 

1 . Determine mentally, if possible, some number a for which 
the polynomial becomes zero. (Factor theorem.) 

2 . Divide the polynomial by (x - a) by synthetic division. 
This division will give further assurance that the remainder 
is zero, and will also determine the other factor, — the quotient. 
The quotient often may be factored. 

Example. — Find the factors of rr 3 + 3 x 2 - 4 x - 12. 

Solution. — 1. When x = 1, then (mentally) x 3 + 3x 2 -4x-12 
— ~ 12- a; — 1 is not a factor of the polynomial. 

2. When re = - 1, z 3 + 3 z 2 - 4 z - 12 = - 6. /. x + 1 is not 
a factor. 

3. When rc = 2, re 3 + 3 a; 2 — 4 x — 12 = 0. .*. rr - 2 is a factor. 

4. Dividing by synthetic division: 

£ + 21 x 3 + 3 rc 2 - 4 a; — 12 
| 1 +2 +10 +12 

1+5 + 6 || 0 

Remainder =0. .'. £—2 is a factor. The other factor is £ 2 + 5 x + 6. 

5. re 3 + 3 £ 2 - 4 £ - 12 = (x - 2)(£ 2 + 5 x + 6) 

= (£ “ 2 )(£ + 2)(x + 3 ). 



SPECIAL PRODUCTS AND FACTORING 303 


EXERCISE 149 


Find the factors of: 

1. a?-4:X 2 + x + 6 

2. y 3 + 2 y 2 - 9 y - 18 

3. m 3 -5m 2 + 3m-f9 

4. r 3 -f 3 r 2 — 4 r — 12 


6 . s 3 - 5 s 2 - 2 5 + 24 

6. 2 x 3 + 6 x 2 + 3 x — 2 

7. 3 2/ 3 - 10 ?/ 2 -t- 2 2/ -f- 3 

8 . r 3 —3r 2 -3r—4 


9. s 4 + s 3 — 3 s 2 — 4 5 — 4 

10. ra 4 + 5 ra 3 + 5 m 2 — 5m — 6 

11. Factor x 3 + 4 ax 2 — 3 a 3 


Solution. — 1. Is (x + a) a factor? 




a x 3 + 4 ax 2 + 0 • a 2 x — 3 a 3 

— a _ etc. etc. 

1 + 3 a etc. etc. 


(Complete the solution.) 


12. x 3 - 4 ax 2 + 5 a 2 x - 2 a 3 13. x 3 - 4 tx 2 + t 2 x +6 t 3 


14. x 3 — 4 cx 2 — 15 c 2 x -H 18 c 3 
16. x 3 — 5 mx 2 — 8 m 2 x + 48 m 3 


EXERCISE 150 

Miscellaneous Examples in Factoring, Fractions , 
and Equations Solved by Factoring 

Factor: 


1. a 2 bc — ac 2 d -f ab 2 d — bed? 

2. 4 ax + 5 by — 4 ay — 5 for 

3. 4 a 2 4- 6 5 — 9 — b 2 

4. 16c 2 — 4 d 2 + 20 rd — 25 r 2 
6. 3 a 3 - 21 a 2 - a + 7 

6. x 2 — y 2 — z 2 + 2 2 /z 

7 . y 3 -y 2 -y - 2 
8 ^10 _ r io 

9. x 6 + 2 x 3 + 1 


10 . 4 a 3 + 15 — 6 a — 10 a? 

11 . x 3 - 3x 2 - 10 x + 24 

12. 5 m 5 n — 5 mn 3 

13. c 3 + 2 c 2 - 9 c - 18 

14. 6 t 3 - 9 t + 8 t 2 — 12 
16. m 3 — 13 m — 12 

16. x 4 - 2 x 2 i/ 2 + t/ 4 

17. x 4 — x 3 — 5 x 2 — x — 6 

18. ax 3 + fex 2 — ac 2 x — fee 2 





304 


ALGEBRA 


19. a 3 - 19 a - 30 22. r 3 + 2 - r * 2 - 3 r 

20. 5 rx + 5 ry — sx — sy 23. a 3 6 3 — 8 6 3 — 8 z/ 3 -f- a 3 ?/ 3 

21. a 6 - a 4 - a 2 + 1 24. 2 c 4 - 4 c 3 - 3 c 2 + 6 c 

25. (a 2 + 2 a) 2 - 11 (a 2 + 2 a) - 24 
26. z 4 + 2 x?y 2 + 9 i/ 4 27. z 6 + 7 z 3 — 8 

28. a 3 + 2 a 2 6 — 5 ab 2 — 6 6 3 

29. 4 m 3 - 4 m 2 ic + 7 raz 2 - 7 x? 

30. y-jjr-g — 


Reduce to lowest terms : 

31 x? -y 2 - 2 yz - z 2 
x 2 + 2 + y 2 — 2 2 

a 3 - 3 a 2 + a - 3 
a 3 — 2 a 2 + a — 2 


c z -3 Jd + cd 2 - 3d* 

c 2 -9<? 

a 3 + 2 s 2 y - xy 2 - 2 y 3 

a: 2 - 4 y 2 


Find: 

a 2 — 2 a6 + b 2 - c 2 , a 2 + 2 ab + 6 2 — c 2 
a 2 — 6 2 — 2 be — c 2 a 2 — 6 2 + 2 be — c 2 


ax — ay + by — bx < aa; -j~ ay ~ fa ~ fry 
x 2 — y 2 a 2 — 2 ab + 6 2 

m 3 -f 2 m 2 n — mn 2 — 2 n 3 _ m 4 — 2 m 2 n 2 + 

m 2 - 4n 2 m 3 — 2 m 2 n + mn 2 — 2 n 3 

a^ — 3 a: + 2 a; 3 - a; 2 + 2 g - 2 

38 ‘ 3 a: 3 + 6 a; 2 + x + 2 3 a; 4 + 7 a: 2 + 2 


39. 


2 rt — 2 rn — 3 st + 3 sn 

rt — rn — sn + st 


r 3 — rs 2 + r 2 s — a 3 
2 r 2 — rs — 3 s 2 


40. 



ab — c 2 \ 
a 2 + ab + b 2 ) \ 


3 ab + c 2 \ 
a 2 + a6 + b 2 ) 


41. Solve the equation x 2 + ax — 3 ab — 3 bx = 0. 
Solution. — 1. x 2 + (xx — 3 ab — 3 bx = 0. 

2. Factoring, x(x + a) — 3 b(x + a) = 0. 

3. (x — 3 b)(x + a) = 0. 

4. xi = 3 b, and x% =• — o. 


(See page 147) 


















SPECIAL PRODUCTS AND FACTORING 305 


Check. — When * = 3 6, does (3 6) 2 + o(3 b) - 3 ab - 3 6(3 6) = 0? 

Does 9 6 2 + 3 ab - 3 ab - 9 6 2 = 0? Yes. 

Similarly for the root x = — a. 


Solve the following equations for x : 

42. x 2 — ax bx — ab — 0 

43. 6 x 2 + 2 nx — 3 mx — mn = 0 

44. x? - 4 x + 3 x 2 - 12 = 0 

46. x 3 — 3 x — 2 x 2 -f- 6 = 0 

46. ax 2 — oca: — 6x -f 6c = 0 

47. abx 2 — 2 6x -|- 3 ax — 6=0 

48. x 2 — r 2 — 2 rs — s 2 = 0 

49. x 2 — to 2 + 2 mn — n 2 = 0 

50. x 2 — bx — cx — ax + a6 + ac = 0 

51. a 2 -ox=2x + a + 6 

52. acx 2 + bcx — abx — 6 2 = 0 

53. 2 mnx 2 + nx — 2 mx — 1=0 

54. (fix 2 — b 2 -f 2 be — c 2 = 0 


55. edmx 2 — amx — ednx + an = 0 


x 2 — x + 1 


66 . 


67. 


60. 


x + 1 


S + 


1 


9 x 2 

y 2 


X + 1 
8j/ 

3 x 


+ i + 

2 y 3 x 


58. 


59. 




(i+-V - — 

Vy x/ xy 

£ , §J/ 

</ 2 x 2 

16 m 2 - 

_ m z 

0 + ri?) “ i 


xy 


y 3 


x 2 + 2 xy + 2/ 2 / ar* + y z 


201. Equations of higher degree solved by factoring. A 

first-degree equation, page 104 (having one unknown), has 
one root; a second-degree equation has two roots, page 148. 











306 


ALGEBRA 


It is proved in higher courses in algebra that an equation of 
degree n has n roots. 

The roots of equations of degree higher than the second are 
not obtained readily except in special equations which can be 
solved partially at least by the factoring method. 

Example 1. — Find the roots of x 4 — 13 x 2 + 36 = 0. 
Solution. — 1. Factoring, (x 2 — 4)(x 2 — 9) = 0. 

2. .*. x 2 - 4 = 0; /. x 2 = 4; /. * = 2, or x = - 2. 

Also, x 2 — 9 = 0; x 2 = 9; x = 3, or x = — 3. 

Example 2. — Solve the equation y z -f 2 y 2 — 4 y + 1 =0. 
Solution. — 1. Factoring by the factor theorem, 

C V ~ D(y 2 + 3 y - 1) = 0. 

2. y — 1 = 0, or y = 1. 

3. Also, y 2 + 3 y — 1 = 0; y = ~3± v 9 + * or ~3± VjE . 

z z 

4. /. y = - 3 ± 3 605 ; y = .302+ or - 3.302+ 

5. Hence, yi = 1; y 2 = .302+; y, = — 3.302 + . 


EXERCISE 181 


Solve the following equations : 

1 . x* — 5 ^ + 4 = 0 

2 . y 4 -29 y 2 + 100 = 0 

3. a 4 - 8 z 2 - 9= 0 

4. w 4 — 5 jo 2 — 36 = 0 
6. i 4 + 12 < 2 - 64 = 0 

6. 4 x 4 — 13 x 2 + 9 = 0 

7. 4 y 4 — 19 y 2 + 12 = 0 

8 . 9 a 4 - 35 n 2 - 4 = 0 


9. 6a: 4 -5^ + l= 0 

10. 2 z 4 + 11 x 2 + 5 = 0 

11. x? - x? - 5 x + 5 = 0 

12. x 3 -2x 2 -z + 2= 0 

13. y 3 —2y 2 — 6y—3 = 0 

14. 2y 3 + y 2 -6y+3=0 
16. c 3 -6c 2 + llc-6=0 
16. 2 c 3 -5c 2 + c + 2= 0 
41 + 4=0 


17. 61 3 — 11 f 

18. ! 4 + l 3 - 2 < 2 - t + 1 = 0 

19. a 4 — 5 a 3 + 5 a 2 + 5 a — 6 = 0 






SPECIAL PRODUCTS AND FACTORING 307 


20. x A — z 3 - 7 z 2 + x + 6 = 0 

21. x A — n? — 4 x 2 + 2 x + 4 = 0 

22. f + 3 y 2 ~ 4 = 0 

23. z 4 — z 3 — 3 z 2 - 3 z — 18 = 0 

24. 6 t A + 5 1? - 5 t 2 - 5 t - 1 =0 

202. Graphical solution of equations. In § 201, the state¬ 
ment was made that an equation of the nth degree, having one 
unknown, has n roots, but that these roots are not readily 
found for equations of degree above the second. Such equa¬ 
tions may be solved graphically as in § 188. 

Example. — Solve the equation x? — 4x 2 — 2 x + 8 =0. 

Solution. — 1. Let ?/ = x 3 — 4x 2 — 2 « + 8. 


When x = 

0 

1 

2 

3 

4 

5 

- 1 

- 2 

- 3 

then y = 

8 

3 

- 4 

- 7 

0 

23 

5 

- 12 

- 49 



2. The curve crosses the horizontal axis at points A, B , and C. 
Hence its roots are, approximately, — 1.42, + 1.42, and + 4. 



















































































































308 


ALGEBRA 


Check. — This equation can be solved as in § 201. Using the factor 
theorem: z 3 — 4 x 2 — 2 x + 8 = (x — 4) (x 2 — 2) = 0. 

x = 4; also x 2 = 2, or x = ± v^2 = ± 1.414. 

Clearly the results ±1.42 obtained graphically are close to the roots 
± 1.414. 

EXERCISE 152 

Solve the following equations graphically: 

1. z 3 - 3 z 2 - z + 3 = 0 6. a?-x*-8x + 8= 0 

2. x 3 — 4 x 2 — 7 x + 10 = 0 7. x 3 — 4 x 2 — x + 4 = 0 

3. a? + x 2 — 6 x = 0 8. x 3 x 2 — 5a; + 3 = 0 

4. x 4 — 10 z 2 + 9 = 0 9. z 3 + 2 x 2 + x + 2 = 0 

5. z 3 + x 2 - 10 x - 10 = 0 10. z 4 - z 3 - 8a: 2 + 12a; = 0 


XVI. QUADRATIC EQUATIONS HAVING TWO 
VARIABLES. GRAPHICAL SOLUTION 


203. Graph of a single equation. 

Example 1 . — Draw the graph of y — x 2 = 0. 
Solution. — 1. Solve the equation for y in terms of x. 


If X 

0 

1 

2 

3 

4 

5 

- 1 

- 2 

- 3 

- 4 

- 5 

then y = 

0 

1 

4 

9 

16 

25 

+ 1 

+ 4 

+ 9 

+ 16 

+ 25 


3. The graph obtained is a 

parabola. 

4. The coordinates of any 
point on the graph satisfy the 
equation. 

Thus, A = (- 4.5, 20+). 
Substituting in y = x 2 , does 
(20+) - (- 4.5) 2 = 0? 

Does (20+) - (20.25) = 0? 
Approximately, yes. 

5. The coordinates of any 
point not on the graph do not 
satisfy the equation of the graph. 
E.g., you will find that the co¬ 
ordinates of the point (3, 15) do 
not satisfy the equati'on. 



Example 2. — Draw the 

graph of x 2 + y 2 = 25. _ 

Solution. — 1. Solve the equation for y. y = ± ^25 — x 2 . 


2 . 


When x = 

0 

+ 1 

+ 2 

+ 3 

+ 4 

+ 5 

then y = 

± V25 
± 5 

± V24 
± 4.8 

± V21 

± 4.5 

± vTe 

± 4 

± 

± 3 

K- 

0 ol 


309 




































































































310 


ALGEBRA 


3. When ar is — 3, y = ± V25 — (— 3) 2 = ± V25 — 9 = ± 4. 
Similarly, when x = — 1, — 2, — 4, and — 5. 

4. y has two values for each value of x. Thus, when x = + 4, 
y = +3, or — 3. Therefore (4, 3) and (4, — 3) are points on the 
graph. 

5. When x is greater than 
5 , y is imaginary. Thus, 
when x = 6, y =±v / 25—36 
or ± V— 11. This means 
that there are not any points 
on the graph for values of x 
greater than + 5, or less 
than — 5. 

6. The graph appears at 
the right. 

It is a circle. 

7. Every equation of the 
form x 2 + y 2 4 — r 2 is a circle 
with its center at the origin, 
and its radius equal to r. 

Example 3 . — Draw the graph of 9 .r 2 + 25 y 1 = 225 . 

Solution. — 1. y 2 * * = — ~ 9 ^ y = ± | V225 - 9 z 2 . 


2. When x = 2, y = ± ^V225 - 36 = ± £^189 = ± i(3V2T) 
= ± i(4.5) = ± 2.7. Similarly, 


When 

x = 

0 

+ 1 

+ 2 

+ 3 

+ 4 

+5 

+ 6 

then 

y = 

± V225 
5 

±iV2l6 

±£\ / 189 

±iVl44 


Vo 

± iV- 99 
• / 

— 

± V- 

± *(14.6) 

± M13.5) 

±¥ 

±'t 

0 

imag’y 


±3 

± 2.9 

± 2.7 

± 2.4 

± 1.8 

0 

imag’y 


3. For x = — 3, y has the same values that it has for x = + 3. 
Similarly when x= — 1, — 2, — 4, and — 5. 

4. For each value of x there are two values of y. The graph follows. 

5. The graph is an ellipse. Every equation of the form ax 2 + by 2 

= c, in which a, b, and c are positive, and a does not equal b , has for 

its graph an ellipse. 



























































































EQUATIONS HAVING TWO VARIABLES 


311 



Example 4. — Draw the graph of 9 x 2 — 4 y 2 = 36. 


Solution. — 1. y 2 = ® X " \ —— ; V — ± f Vx 2 — 4. 

2. When x = 1, y = ± - 4 = ± /. V is imaginary. 


When 

x = 

0 

+ 1 

+ 2 

+ 3 

+ 4 

+ 5 

+ 6 

then 

y = 

± 

imag’y 

± W^3 
imag’y 

± fVo 
0 

± fVs 
± 3.3 

± fVl2 
± 5.1 

± fV21 
± 6.8 

±|V32 
± 8.4 


























































































































































































































































312 


ALGEBRA 


3. When x = — 1 , y = ± fV(— l) 2 — 4 = db f Vl — 4 = ± f V^~3. 
These are the same values obtained when x = + 1. Similarly for 

x = — 2, — 3, — 4, etc. The graph follows. 

4. This graph is a hyperbola. Every equation of the form ax 2 — 
by 2 = c, where a, b, and c are positive members, is a hyperbola. 


EXERCISE 153 

Draw the graphs of the following equations. Name the 
curves obtained. 

1. x 2 + y 2 = 49 


2. y = 6 x 2 

3. x = 4 y 2 

4. x 2 + 9 y 2 = 81 

5. x 2 — 9 y 2 = 16 


6. x 2 + y 2 = 70 

7. xy = 4 

8. xy = — 4 

9. 4 x 2 + 2/ 2 « 36 
10. 9 x 2 + 4 ?/ 2 = 36 


11. Recalling the remark in Step 7, page 310, draw the 
following graphs, without any computation. 


a. x 2 + y 2 = 36 


b. X 2 + y 2 = 81 


c. x 2 + y 2 = 121 


SUPPLEMENTARY EXAMPLES 

12. Draw the graph of x 2 — xy + y 2 = 63. 

Solution. — 1. Another method of obtaining points on a graph 
follows: 

Let x = 0; then y 2 = 63, or y = ± V 63, or ± 7.9. 

.*. x = 0, y = + 7.9 is a 1st point on the graph, and x = 0, y = 
— 7.9 is a 2d point. 

2. Let x — — 1. .*. 1 + y + y 1 — 63, or y 2 + y — 62 = 0. 

• v = ~ 1 ± V l + 248 _ - 1 ± V249 _ - 1 ± 15.7 
y 2 2 2 
or 7.3; and y 2 = - = — 8.8. 

That is, x = — 1, y = 7.3 is a 3d point, and x = — 1, y — — 8.8 
is a 4th point on the graph. 

In this manner, complete the table below and draw the graph. 


When x = 

0 

- 1 

- 3 

- 6 

- 9 

+ 3 

+ 6 

+ 9 

then y\ = 
and y 2 = 

+ 7.9 
-7.9 

+ 7.3 
-8.8 

























EQUATIONS HAVING TWO VARIABLES 


313 


13. y 2 + x 2 — 6 x — 16 = 0 15. x 2 + y 2 — 8 x — 4 y = 5 

14. y 2 + 4 y + x 2 — 32 =0 16. y 2 — xy — x 2 = 19 

17. On the same set of axes draw the graph of xy = 5, and 
of xy = — 5. Notice the effect of changing the sign of 5. 

18. Draw on one set of axes y = x 2 and y = — x 2 . 

19. Draw on one set of axes x 2 + y 2 = 25 and x 2 — y 2 = 25. 

20. Draw on one set of axes : 

a. X? + y 2 = 100; b. x 2 + 4 y 2 = 100, and c. 4 x 2 + y 2 = 100. 


Example 1. — Solve the system 




204. Solution of simultaneous quadratic equations graphically. 

+ y 2 = 25 (l) 

y + 1=0 (2) 

Solution. — 1. In Example 2, page 310, the graph of x 2 + y 2 = 25 
is found to be the circle of radius 5, with center at the origin. 

2. For equation 2: when x — 0, - j/ + 1 = 0, or y = 1. 

When x = 2, 2 - y + l = 0, or y = 3. The graph is the straight 



3. A is on both graphs. Its coordinates are (3, 4). By substitu¬ 
tion, you will find x = 3, y = 4 satisfy both equations. 

4. Similarly the coordinates (-4, — 3) of B satisfy both equations. 

Note. — The graph of every linear equation having two variables is 
a straight line. (Rule, page 218.) The graph of every quadratic equa¬ 
tion having two variables is one of the four curves discussed in § 203. 












































































314 


ALGEBRA 


Since a straight line will, in general, cut such a curve in two points, a 
system consisting of a linear equation and a quadratic equation will 
have two common solutions. 

If the straight line touches the curve at only one point (is tangent 
to the curve), the system has one “double” solution. If the straight 
line fails to touch the curve, the system has two imaginary solutions. 

Example 2. — Solve the system { ,r y ~ ^ 

l x 2 + 2 y z = 34 (2) 

Solution .— 1. The graph of equation (1) is the circle obtained on 
page 310. The graph of the equation (2) is the ellipse in the figure 
below. (It is obtained as on page 310.) 



2. The common points on the graphs are: 



By substituting these values in equations (1) and (2), you will find 
that these are common solutions of the given system of equations. 

Generally, two quadratic equations have four common solutions, just 
as do these two equations. This becomes clear, if you draw upon one 
set of axes any two of the four curves found in § 203. For example, 
























































































































EQUATIONS HAVING TWO VARIABLES 


315 



However there are other possibilities. For example, the ellipse may 
intersect only one branch of the hyperbola. Then the system has two 
real solutions and two imaginary solutions. 


EXERCISE 154 


Solve the following systems graphically: 

x 2 - 4 y = 20 
x 2 y 2 = 25 
x 2 -y 2 = 19 
2 x + y 2 = 5 
4 z 2 + 7 y 2 = 32 
lly 2 - Sx 2 = 41 
x 2 — A. y 2 = 28 
— x = 4 : l 2 X - 3 y = 7 


z 2 + y 2 = 52 

5. 

x - y = 2 


4 x 2 + y 2 = 40 
x + y = 5 

6. 

x 2 + y 2 = 34 

7. 

xy = 15 


xy = 45 

8. 


9. a. Draw the graph of x 2 + y 2 =25. 

b. On the same axes, draw the graph of y = x 2 , and obtain 
the common solutions of it and x 2 -f y 2 = 25. 

c. Repeat part b, using y = x 2 + 5. 

d. Repeat part b, using y = z 2 - 8. 

e. Repeat part b, using y = x 2 + 8. 

















































































































XVII. SYSTEMS OF EQUATIONS INVOLVING 
QUADRATICS 

205. Review the definitions of independent equations (§ 150); 
simultaneous equations (§ 150); system of equations (§ 150); 
and dependent equations (§ 155). 

206. Solving a system consisting of one linear and one 
quadratic equation. 

Example. — Solve the system { 3 c 4- 2 d — 2 (1) 

* \cd + 8 c = 4 (2) 

Solution. —1. From (1), c = 2 T - 2 d . (3) 

O 

2. Substituting in (2), — — — —^ + 8 ^^ = 4. (4) 

3. Simplifying, — 2 d — 2 d 2 — 16 — 16 d — 12 = 0. (5) 

4. 2 d? + 18 d + 28 = 0, or d 2 + 9 d + 14 = 0. 

5. Solving for d, d = — 2, or d = — 7. 



When d = — 2, 

When d = — 7, 

6. 

in (1), 3 c - 4 = - 2 

in (1), 3 c - 14 = - 2 


or c = f. 

or c = 4. 


7. .'. The solutions are a. I c ® b. I c ^ 

\d = - 2 \d = - 7 

These solutions check when substituted in (1) and (2). 


A system consisting of one linear and one quadratic equation 
always has two solutions. 


EXERCISE 155 


Solve the following systems of equations: 


\x 2 + y 2 = 85 
\x - y = 1 
| 4 r 2 + t 2 = 25 


3 f3a — 2b = 3 

‘ j 3 a 2 - 4 ab + 2 b 2 = 27 

4 | z 2 - 3 xy + y 2 = - 11 
i y - x = - l 


316 







EQUATIONS INVOLVING QUADRATICS 317 



x 2 + y 2 — x — 2y = 0 
x +2y = 5 
2 c — d = 1 

4 cd + d?—3 c 2 = 9c — 7d 


17. 





207. Quadratic equations homogeneous except for the con¬ 
stant term. 


Examples. — 4 z 2 - xy + 3 y 2 = 5; | x 2 - 1 1 / 2 + 2 xy = f 


Observe: 

a. The variables do not appear in a denominator. Therefore 
the equation is integral. 

b. The variables do not appear under radical signs. There¬ 
fore the equation is rational. 

c. The sum of the exponents in the terms which do contain 
the variables is 2. It is this characteristic which makes the 
equation homogeneous of degree two. 

Some systems consisting of two quadratics, homogeneous 
except for the constant term, can be solved by eliminating 
the constant terms. 


318 


ALGEBRA 


Example. 

ci + f a 2 + ab + b 2 = 63 

— Solve the system j ^ _ ft2 = _ ^ 

(1) 

(2) 

Solution. — 

- 1 . To eliminate 63 and 27: 


M 3 (1) 

3 a 2 + 3 a 6 + 3 6 2 = 189. 

(3) 

M 7 (2) 

7 a 2 - 7 6 2 = - 189. 

(4) 

(3) + (4) 

10 a 2 + 3 ah - 4 6 2 = 0. 

(5) 

2 . Solve (5) for a in terms of 6 : 



(5 a -f- 4 6 ) (2 a — 6 ) = 0. 5 a + 4 6 = 0; and 2 a — 6 = 0. 


3. Solve the two systems below: 

r a 2 - 6 2 = - 27 
\ 5 a + 4 6 = 0 
„ _ 46 

5" 

1 ®_ 6 2 - = - 27. 

25 

16 6 2 - 25 b 2 = - 675. 

- 9 6 2 = - 675. 
ft 2 = 75. 

6 = ± V 75 , or ± 5^3. 

But a = - 

5 

a — 5 v ^ 3 ). 

a = T 4V3. 

Two solutions are : 


fa 2 - 6 2 = - 27 
\ 2 a - b = 0 
.’. 6 = 2 a. 

a 2 _ 4 a 2 = _ 27 . 

- 3 a 2 = - 27. 

/. a 2 = 9. 

.'. a — ± 3. 

.’. 6 = 2(± 3), or ± 6 . 

/. Two more solutions are: 

n f a = +3 n j a = — 3 
\ 6=+6 \ 6=-6 




- 4V3 
+ 5V3 




+ 4 V 3 
- 5V3 


Check. — These solutions may be checked by substituting them in 
the original equations. 

Note 1 . — In the first column of Step 3, notice that the value — 4 V3 
of a comes from the value + 5 V 3 of 6 . Therefore they form a solu¬ 
tion (Solution A). 

Note 2. — If one equation does not have a constant term, solve it 
at once for one variable in terms of the other, as in Step 2 above. 

A system consisting of two quadratics has four solutions, — 
of which some may be imaginary. 



V 


EQUATIONS INVOLVING QUADRATICS 

EXERCISE 156 


319 


Solve the following systems: 


'■1 

: z 2 + 3 xy = 28 
[ xy + 4 y 2 = 8 

7. 

2. < 

[ a 2 + y 2 = 37 
[xy = 6 

8. 

s -l 

f m 2 + mn = 10 
[ mn + n 2 = 15 

9. 

M 

a 2 — 3 ab + 2 b 2 =3 

2 a 2 + b 2 = 6 

10. 

6. 

| 3 xy - 2 y 2 = 0 
\ 2 x 2 — xy = 2 

11. 

6. 

f 3 x 2 - 5 xy + 3 y 2 = 15 
[ .r 2 — xy = 4 

12. 


f 3 cd + cP = - 14 
\ c 2 — cd = 30 
| x 2 — xy = 4 

l z 2 + y 2 = io 

| 4 z 2 + 3 y 2 = 28 
1 2 x 2 + a:?/ + y 2 = 8 
j 2 rt - 2 r 2 - t 2 = - 2 
1 * 2 - r 2 = 3 
J 2 a ; 2 — 3 xy = 2 
\ 4 z 2 -f 9y 2 = 10 
j h 2 + 25 r 2 = 125 
l 5 r 2 + rh = 15 


208. The number of common solutions of a system of equa¬ 
tions. In § 206, where one equation is linear and the other is 
quadratic, each system has two common solutions. In § 207, 
where both equations are quadratic, each system has four 
common solutions. The number agrees in both cases with the 
number determined by the following 


Rule. — Two rational and integral equations having two 
variables, whose degrees are m and n respectively, have mn 
common solutions. 

Thus, a system consisting of a cubic equation and a quadratic equa¬ 
tion should have 3 • 2 or 6 common solutions. Some of these may 
be imaginary, and some may be repeated solutions. For example, 
x = 2; y = 1 may appear twice in the list. 

209. Equivalent systems of equations are two different 
systems which have the same common solutions. In terms of 
graphs, the intersection points of one system are the same as 
those of the other system. 


320 


ALGEBRA 


Example. — Solve the system 


a 


- y 
2 + y 1 


210. Solution of systems which are reducible by division to 

equivalent systems whose equations are of lower degree. 

' 4 = 240 (1) 

= 20 (2) 

Solution. — 1 . By the rule of §208, it would appear that there 
should be 4 • 2 or 8 common solutions. 

In this example, however, x 4 — y 4 can be divided by x 2 + y 2 . 

2 . Dividing (1) by (2), 

3. Now form the system 

This system will have 2 • 2 or 4 common solutions. 

4. (3) +(2): 2 x 2 = 32; x 2 = 16; x = ± 4. 

5. When x = + 4, 16 + y 2 = 20; y 2 = 4; y = =*= 2. 

.*. x = 4, y = 2, and x = 4, y = — 2 are two solutions. 

6 . When x = — 4, 16 + y 2 = 20; y 2 = 4; y = =*= 2. 

x = — 4, y = 2, and x = — 4, y = — 2 are two more solutions. 

Check all four solutions by substituting in equations ( 1 ) and ( 2 ). 
Note. — Whenever possible, divide one equation of the given 
system by the other, member by member, and form a new system 
consisting of the quotient equation and the divisor equation. 


II 

1 

(3) 

f x 2 — y 2 = 12 

(3) 

{ x 2 + y 2 = 20 

( 2 ) 


EXERCISE 157 

Solve the following systems of equations: 


■•I 

' x 2 — y 2 = 14 

6. , 

fa 2 — 6 2 =4a + 66 — 

. x + y = 7 

[ a — b = 2 

1 i 

f x 4 — y 4 = 65 

7 ' i 

f x 3 — y 3 = 218 

CO 

r - 1 

II 

+ 

CM 

[ x 2 + xy + y 2 = 109 

3. 1 

O 

iO 

II 

1 

8 - | 

| x 3 — x 2 y = 54 

i y(* - y) =25 

l x - y = 6 

4 'i 

f 2 a? + 11 xy + 12 f = 63 


05 

i —i 

II 

CO 

1 

T* 

{2x + Zy = 9 

1 x 2 + xy + y 2 = 19 a 2 

6. J 

1 

[ z 2 + 2 xy — 3 y* = 17 

10 -| 

[ x? - 8 y 3 = 208 

L* +3t/ = 17 

t-H 

II 

1 


11 . 


x 3 + y 3 = 35 
x+y = 5 


EQUATIONS INVOLVING QUADRATICS 


321 


13. 

14. 


12 . 


{: 


+ 6 3 = 144 
+ 6=4 

z 2 + xy — 6 y 2 = 21 
xy + 3 y 2 = 84 


£ 3 — y 3 — 56 x + 28 y — 56 

z - 2 /= 2 z + ?/-2 

+ ab 3 


” {£ 


16 


84 

a 3 6 + 6 4 = 28 
f a ; 2 — 2 xy + 6 x 
1 xy - 2 y 2 + 6 y 


= 32 

= 8 


EXERCISE 158 

1 . Find two numbers whose sum is — 9 and the sum of 
whose squares is 101 . 

2 . The sum of the squares of two numbers is 1274 and the 
larger is 5 times the smaller. Find the numbers. 

3 . The product of two numbers is 5 and the sum of their 
squares is 26. Find the numbers. 

4 . Find two numbers whose sum is 10 and whose product is 9. 

6 . Find two numbers whose difference is 5 and whose 

product is 6 . 

6 . The product of two numbers is 48 and the sum of their 
squares is 100 . Find the numbers. 

7 . The sum of two numbers is 8 ; the square of the first 
number increased by the product of the two numbers is equal 
to 49 diminished by the square of the second number. Find 
the numbers. 

8 . The square of the length of the diagonal of a rectangle 
is 40 and the area of the rectangle is 12 square feet. Find the 
dimensions of the rectangle. 

9 . Find two numbers such that the square of the first 
diminished by the product of the two numbers shall be 15, 
and such that the difference between their squares shall be 21 . 

10 . Find two numbers such that their product increased by 
the first number shall be 8 , and such that their product in¬ 
creased by twice the second number shall be 12 . 


322 


ALGEBRA 


11. In 6 hours, two boys row 16 miles downstream and back 
again. Their rate upstream is twice the rate of the current. 
Find the rate at which they row in still water and the rate of 
the current. 

12. The sum of the squares of the two digits of a number is 
74. If 18 be added to the number, the digits of the sum are 
the digits of the original number in reverse order. Find the 
number. 

13. Find the sides of a rectangle if the area of the rectangle 
increased by the shorter side is 15 and if the area increased by 
the longer side is 16. 

14. The sum of the volumes of two cubes is 72; and an edge 
of one cube plus an edge of the other cube is 6. Find the length 
of the edges of each cube. 

15. Find the number of two digits such that if the digits be 
reversed, the new number minus the original number is 18, 
and their product is 403. 

16. Find two numbers the sum of whose squares increased 
by the product of the numbers is 28, and the difference of whose 
cubes is 56. 

17. The area of a trapezoid is 36 square inches. Its altitude 
is 3 inches. Find the two bases of the trapezoid if the square 
of the length of the lower base is 192 more than the square of 
the length of the upper base. 

18. The sum of a certain fraction and its reciprocal is 

If three times the numerator is decreased by twice the de¬ 
nominator the result is equal to -12. Find the fraction. 

19. In an isosceles triangle, the vertex angle is 60° more 
than the square of either base angle. Find the angles of the 
triangle. 

20. The sum of the squares of the two digits of a number is 
20. If the number itself is 4 times the sum of its digits, find 
the number. 


EQUATIONS INVOLVING QUADRATICS 323 

21. Find the number of two digits in which the units' digit 
exceeds the tens', digit by 2, and such that twice the number 
exceeds the sum of the squares of the digits by 40. 

22. The diagonal of a rectangle is 13. If the rectangle were 
4 inches shorter and 2 inches wider, the diagonal would be 
Vll3. What are the dimensions of the rectangle? 

23. Two boys on bicycles make a trip of 15 miles. The 
sum of their rates is 15 miles per hour. The first boy requires 
50 minutes more for the trip than the second boy. What are 
their rates? 

24. The area of a certain rectangular field is 260 rods. Its 
perimeter is 66 rods. What are its dimensions ? 

25. The area of a certain garden is 1000 square feet. If 
the length be decreased by 5 feet, and the width be increased 
by 5 feet, the area is increased by 50 square feet. What are the 
dimensions of the garden? 

26. The area of a certain triangle is 108 square feet. Its 
base exceeds its altitude by 6 feet. What are the base and 
altitude of the triangle? 

27. If the larger of certain two numbers be multiplied by 
the difference of the numbers, the result is 44. The product 
of the two numbers is 77. What are the numbers ? 


SUPPLEMENTARY EXAMPLES 


211. Miscellaneous types and methods. Many systems of 
equations which cannot be solved by the methods already given 
may be solved by combining the equations so as to obtain a 
linear equation or an equation of the form xy = a constant. 

Example 1. — Solve the system : 



( 1 ) 

( 2 ) 


324 


ALGEBRA 


Solution. — 1. M 2 (2) 2 xy = 12. (3) 

2. Adding (1) and (3), x 2 + 2 xy + y 2 + 2 x + 2 y = 35. (4) 

/. (x + y) 2 + 2(s + y) - 35 = 0. 

(* + y + 7)(x + y - 5) = 0. (§ 205) 

x -f- y = — 7, or x + y = 5. (§ 110) (5) 

3. Form the systems: A \ X ^ y ~ ^ B [ x + y ~ % 

J [ xy = 6 \ xy = 6 

4. Solving A, x = — 1, y = — 6; or, x = — 6, y = — 1. 

Solving £, x = 3, y = 2; or, x = 2, y = 3. 

Check. — The four solutions check when substituted in equations 
(1) and (2). 


Example 2. — Solve the system 


Solution. 


m -f n = 5 + ww 

1. Square (2), 
m 2 + 2 mn + n 2 = 25 + 10 mn -f- m 2 n 2 . 
2. Subtract (1) from (3), mn =18 + 10 mn + m 2 n 2 . 
m 2 n 2 + 9 mn + 18 = 0. 

(mn + 6) (mn + 3) = 0; .*. mn = — 6, or mn = 

A jm+n = 5+ mn 


3. Form the systems, 


4. Solving A, 
Solving B, 


- 3. 


mn = — 6 
m + n = 5 + mn 
\ mn = — 3 

m = 2, n = — 3; or, m = — 3, n = 2. 
m = 3, n = — 1; or, m = — 1, n = 3. 


( 1 ) 

( 2 ) 

(3) 

(4) 

(5) 
(§ 107) 


Check. — The four solutions check when substituted in equations 
(1) and (2). 


EXERCISE 159 


a 2 + b 2 = 50 
ab + 7 =0 
x 2 — xy + y 2 = 7 
z + y = 4 

2 a 2 - 3 ab = 12 

=20 

3 x 2 + 8 y 2 = 14 
x 2 - 12 y 2 = 1 


cd — 4 d? = 1 
3 cd — c 2 = — 10 
10 x 2 + 11 xy + y 2 = 324 
x + y = 3(y - x) 
x 2 + xy = 400 
x - y = 7 
x 2 — y 2 = 9 
5y = 3 x 



EQUATIONS INVOLVING QUADRATICS 325 


„ I x 2 + y 2 = 16 + 3 x 
\x 2 - y 2 = 4 

10 ! x 2 - xy + y 2 = 63 
' \* + y = 12 


11 . 


12 . 


a 2 — b 2 = 16 

a 2 + 2 6 2 = 64 
9 — t/ 2 = T6 


i/ 2 - 14 = 3 x 
13 Ja 2 - ab + b 2 = 21 


\2ab-b 2 = 


15 


fed + 4 d 2 =8 
Ic 2 + 3 cd = 28 


15. 


2 a 2 — a6 + 6 2 =30 

a 2 = 2 ab — b 2 

I p(l + .03 r) =840 
\p(l + .07 r) = 960 
(4 2 £ 2 - 3 AB = 88 

17 ' U - B = 6 

1R I 2 * 2 d - * = - 7 
\4 xV + x 2 = 37 

19 I* 1 + 2/ 2 - 25 

19 ' \(* + l) 2 + (y- 0* = 
fx 3 +27J/ 3 = 756 

20 ' {x 2 — 3 xp + 9 i/ 2 = 63 

[ x 2 — xy = 


fx(x + 2 /) = 126 

• \y(x - y) = 20 
„ f 25 to 2 + p 2 =25 

\5 mp = - 12 
8 

04 j* 2 + j/ 2 - ( x + v) = 22 

'lx + t/+xj/ + l= 0 

„ (x 2 + xy + y 2 = 61 
2& - W + * + y - 29 

26 J8x 3 +t/ 3 = 7m 3 

* \2 x + y = w 

.. (2 x 2 — 111/ 2 = 8 

\3x 2 + 13t/ 2 = 248 


28 


f x + y = 2 a 


lx 2 + y 2 = 2 a 2 + 2 
[x 2 + f = 2(« 2 + 6 2 ) 
29 ' U-y -26 

30 fx 2 + 2 / 2 = 16 a 


1 xy + ay 2 = 6 2 


{x 2 + y 2 -2x=8a 

j 7 ® 2 - 10 vt - 12 < 2 = 36 
25 31 ' \* 2 + 6 vt + 4 f 2 = - 4 

32. | x + y “ 6 
[xy = c 

ay 


33. 


f (x + y)(a 2 - b 2 ) = (x - 2 /)(a 2 + b 2 ) 
34 ‘ \ xy = b 2 


2 m + n 


+ 


35. 2 m - w 

I 4 m 2 + n 2 


2 m — n _ 10 

2 m + n 3 
= 45 




XVIII. THE THEORY OF QUADRATIC EQUATIONS 

212. The sum and the product of the roots of a quadratic 
equation bear interesting relations to the coefficients of the 
equation. 

The general quadratic equation is 

V ax 2 + bx -f- c = 0. (1) 

Dividing both members by a, 


x* + ° -x+ c - = 0. 
a a 


( 2 ) 


In § 181, page 271, you proved that the roots of this equation 

(3) 


and = - 6-V6 2 —4oc 
2 a 2 a 


Using these two results : 


I. 

II. 


n + r 2 = 


— 2b 
2a 



_ (-ft) 2 - (V6 2 -4 act 2 = ft 2 . - (ft 2 - 4 ac) 
4 « 2 4 a 2 ' 


(4) 


/. rir 2 


4 ac 
4 a 2 


c 

a 


These relations are expressed by the 

Rules. — In the quadratic equation ax 2 
1. The sum of the roots is — - • 


a 


(5) 


+ bx + c = 0: 

From (4) 


2. The product of the roots is - • 

a 

326 


From (5) 












THE THEORY OF QUADRATIC EQUATIONS 327 


3 . If the coefficient of x 2 is made 1 (as in equation (2) above), 
then the coefficient of x is minus the sum of the roots, and the 
constant term is the product of the roots. (See (2), ( 4 ), (6).) 

Example. — What are the sum and the product of the roots of 
2x 2 = 9ic + 5? 


Solution. — 1. 

2 . 

3. 

4 . 

5. 


2z 2 = 9x + 5. .. 
.*. 2 x 2 — 9 x — 5 = 0. 
.*.0 = 2; b = - 9; c = 




- 5. 
9 . 
2 ’ 


Note. — Always arrange the quadratic as in Step 2, with all terms 
on the left side, and make the coefficient of x 2 be positive. 


EXERCISE 160 


Find by this rule the sum and the product of the roots of the 
following equations. Check Examples 1, 2, 3, and 6 by finding 
the roots. 

1. x 2 +8x + 7= 0 6. 4 o = 7 - 3 o 2 

2. z 2 — 9 2 + 14 = 0 6. x 2 + 5 bx -4 6 2 


3 . 8X 2 — 2x — 15=0 7 . 15 x 2 - a 2 = 2 ox 

4. 63 y 2 - 16 y + 1 = 0 8. 3 x 2 = 4 xt + 2 f 

9. One root of3x 2 -2x-5=0is -1. Find the other 
root. 


Suggestion. — What is the sum of the roots? 

10. One root of 2 x 2 - 7 x + 3 = 0 is 3. Find the other 
root. 

11. One root of 6 z 2 + 1 = 5 x is - Find the other root. 

12. Find k so that one root ofz 2 +4x — & =0 may be — 12. 

Solution. — 1. n + r 2 = - 4. .*. - 12 + r 2 = - 4. /. r 2 = 8. 

2. But rir 2 = — k. — k = (— 12) -8; :. k — ? 

Check. — Substitute in x 2 + 4 x - k = 0 the value of k you find; 
solve the equation, and thus determine whether - 12 really is one root. 



328 


ALGEBRA 


13. Find k so that — f may be a root of 6 x 2 + 7 x + 2 k = 0. 

14. Find k so that f may be a root of 8 a; 2 — 10 a; = — k. 

15. Find a so that 7 may be a root of 7 z 2 — 48 z — a = 0. 

16. Find c so that the roots of 2 z 2 — 8 2 + c = 0 shall be 
equal. 

Solution. — 1. Let the two equal roots be represented by r. 

2. r + r = ? .*. r = ? 

3. But r • r = y .’. c = ? 

Li 

Check. — Substitute in 2 z 2 — 8 2 + c = 0 for c its value. Then 
solve the equation to determine whether the roots are really equal. 

f 17. Find t so that the roots of 3 a; 2 — 4a:-M = 0 shall be 
equal. 

18. Find s so that the roots of 4 x 2 — sx + 9 =0 shall be equal. 

19. Find m so that x 2 — 14 mx + 9=0 shall have equal 
roots. 

20. Find c so that y 2 — 5y — 8c = 0 shall have equal roots. 

21. Find a so that the roots of ax 2 — 3 x + 4 = 0 shall be 
equal. 

213. Forming quadratic equations having given roots. There 
are two methods. 

Example 1.—Form the equation whose roots are 2.4 and 

- .5. 

Solution. — 1. Let the coefficient of x 2 be 1. Then the equation is 
x 2 - (ri + r 2 )x + r x r 2 =0. (§ 212) 

2. But n + r 2 = 2.4 + (— .5) = 1.9; and r x • r 2 = 2.4 • (— .5) 
= - 1.2. 

3. .*. the equation is x 2 — 1.9 x — 1.2 = 0. 

Check. — Either solve the equation, obtaining its roots, or substitute 
the given roots in the equation. 


THE THEORY OF QUADRATIC EQUATIONS 329 


Example 2. — Form the equation whose roots are \ and — f. 

Solution. — 1. If x = |, then x — £ = 0. 

2. If x = — then x + f = 0. 

3. (* - *)(* + \) = 0, or x 2 + i x - | = 0. 

4. .*. 8 x 2 + 2 * - 3 = 0. 

Check as in Example 1. 

Note. — This second method may be used when forming an equa¬ 
tion which shall have three or more given roots. 


EXERCISE 161 


Form the equation whose roots shall be 


1. 

2,5 

8. 

.2, .5 

2 . 

-3,1 

9. 

1.5, - 2.4 

3. 

6, - 1 

10 . 

- 3.6, - .25 

4. 

1 

1 

11 . 

■f 3m, + 2m 

5. 

1 3 

21 T 

12 . 

+ 

1 

6 . 

2 _ 1 

3-> 6 

13. 

\ c, — f c 

7. 

1.5, 2 

14. 

d -\- e, d — e 


is. VE, - VE 

16. 1 + V2, 1 - V2 

17. V3+2, V3 -2 

18. 3 + V6, 3 - VE 

19. b + 2 c, b - 2 c 

20. Vm, — Vm 

21. a -f ^b, a ~ ^ 


DETERMINATION OF THE CHARACTER OF THE ROOTS 

214. Classification of numbers. The numbers considered in 
this text to this point are: 

I. The Real Numbers. 

1. Rational numbers. 

a. The positive and negative integers. 

b. The positive and negative fractions whose terms are 

integers. 

2. Irrational numbers. 

a. Quadratic surds. (§ 170) 

b. Surd expressions, such as 2 + v/ 3- (§ 171) 

II. Imaginary Numbers. 

1. Pure imaginaries, such as V— 3. (§ 1 89) 

2. Complex numbers, such as 2 — V — 3. 





330 


ALGEBRA 


215. Sometimes it is desired to know what kind of roots a 
quadratic has, without any reference to the exact values of 
these roots. 

Thus, the roots of 2 x 2 — 8 x + 3 = 0 are ^ ^ • 

4 

Since 64 — 24 = 40, the roots are real numbers. 

They would be imaginary only if the number under the radical sign 
were negative. 

Since 40 is not a perfect square, the roots are irrational. (See § 170.) 

Since 40 is added to 8 to get one root, and subtracted from 8 to get 
the other root, the roots are unequal. 

Hence the roots are real, irrational, and unequal. 


It is possible, however, to determine the nature of the roots 
without first finding the roots, as in the example just given. 


The roots of ax 2 + bx + c = 0 are, as you know, 


r i = 


— b + Vb 2 — 4 ac 
2 a 


, and 7*2 = 


— b — Vb 2 — 4 ac 
2 a 


Rule. — 1. If b 2 — 4 ac is positive, the roots are real and 
unequal. They are rational if b 2 — 4 ac is a perfect square. 

2. If b 2 — 4 ac = zero, the roots are real, rational, and equal. 
(Why?) 

3. If b 2 - 4 ac is negative, the roots are imaginary. 

Note. — fr 2 — 4 ac is called the discriminant of the quadratic. 
(Why?) 

Example 1. — What kind of roots has 

2 a: 2 — 5x — 18=0? 

Solution. — 1. 6 2 - 4 ac = (- 5) 2 - 4(2)(- 18) 

2. = 25 + 144 = 169 = 13 2 . 

3. .*. by Rule 1, the roots are real, rational, and unequal. 

Example 2. — What kind of roots has 

3 a: 2 + 2 a: + 1 =0? 

Solution. — 1. b 2 — 4 ac = 4 — 4-3-1 

2. = 4 - 12, or - 8. 

3. .*. by Rule 3, the roots are imaginary. 









THE THEORY OF QUADRATIC EQUATIONS 331 


Example 3. — What kind of roots has 
4 x 2 — 12 z + 9= 0? 

Solution. — 1. b 2 — 4 ac = 144 — 4 • 4 • 9 = 144 — 144 — 0. 

2. by Rule 2, the roots are real, rational, and equal. 

Note. — This type of quadratic is most easily understood if the quad¬ 
ratic is solved by factoring. This example becomes (2 x — 3) (2 x — 3) 
= 0. The roots are £ and f. 


EXERCISE 162 


Determine by inspection 

the kind 

of roots of: 

1. 

y 2 - 7 y + 12 = 

0 

7. 

3 x - 

- 9 x 2 + 1 = 0 

2. 

9 £ 2 - 24 x + 16 

= 0 

8 . 

x 2 = 

2x - 7 

3. 

2 m ! - m + 1 = 

0 

9. 

10 = 

4 a; + 5 z 2 

4. 

3 j, 2 = 5 y + 2 


10. 

16 rrv 

o 

II 

1 

5. 

7 x 2 + 5 x = 9 


11. 

9 x 2 • 

- 30 X + 25 = 0 

6 . 

1 + 6 t + 4 < 2 = 

0 

12. 

x 2 - 

4 x + 8 =0 



XIX. LITERAL, FRACTIONAL, AND NEGATIVE 
EXPONENTS 

216. In the preceding chapters, only positive integers have 
been used as exponents. The fundamental definition when m 
is a positive integer is 

a m = a • a • a .... a (m factors) 

217. There are five fundamental laws of exponents. The 

proofs follow, for the case when m and n are positive integers. 

I. Law of exponents for multiplication. 

Just as a 5 X a 7 = a 12 , so a m X a n = a m+n . 


Proof. — 1. a m = a • a • a . a (jn factors) 

2. a n = a • a • a . a (n factors) 

3. a m • a n = {a ■ a . a} {a • a . a} 

(m factors) (n factors) 

4. = a • a . a • a • a . a (m + n) factors 

5. .'. a m • a n = a m+n . § 216 


II. Law of exponents for division. 

Just as a 9 a 4 = a 5 , so a m + a n = a m ~ n (m greater than n). 
p roo f _ i a • a ..... a (m factors) 


Z / UUJ . JL.-------—^-- 

a n ^& (n factors) 

2- — a • a . a (m — n) factors 

3. .'. a m -i- o n = a TO-n . § 216 

III. Law of exponents for finding a power of a power. 

Just as (a 5 ) 3 = a 15 , so (a m ) n = a mn . 

Proof. — 1. (a m ) n = a m -a m -a™ . a m (n factors) 

2 * = a m+m+m .» (n terms) 

3. (a m ) n = a wn . 

332 












LITERAL AND NEGATIVE EXPONENTS 333 


IV. Law of exponents for finding a power of a product. 

Just as ( ab) b = a b b 5 , so ( ab) n = a n b n . 


Proof. — 1. ( ab) n = ( ab ) • ( ab) • ( ab ) ( ab ) (n factors) 

2. ( ab) n = (a • a • a o) (6 • b • b b) 

(n factors) (n factors) 

3. (ab) n = a n • b n . 

V. Law of exponents for finding a power of a quotient. 
t x /a\ 5 a 5 /a\ n a n 

justas (i) = ^ so y-s=- 


Proo/ —!. (f)”=(f) (f) (f) . (f) (n factors) 

/q\ w _ a ' a • a . a (n factors) 

2 ‘ \b) b - b • b .6 (n factors) 


3. 



6 "' 


Involution is the name given to the process of finding a power 
of a number. 


EXERCISE 163 


In the following examples, the literal exponents represent 
positive integers. Find the results of the indicated operations, 
using the five laws above. 


1 . £ 10 * X* 

2. m 6 • to 8 

3. y 4 • y a 

4. z a • x b 

5. ar 2n • x bn 

6. x s • z r 

7. y*- 2 • i/ 3 

8. z 2n • z 5 

9. w n_1 • w n 

10. x 3m ■ x m+1 

11. y 10 2Z 3 


12. z 12 -H z n 

13. t 2n f 3 

14. x 3m -7- a; m 

15. ?/ m+4 - 2/ 3 

16. -5- z m_1 

17. k^ 2 - 

18. w 3r+1 -7- w 2 

19. z n+3 "i" a: 2 

20. a; n-3 -5- a: 2 

21. (ar>) 2 

22. (i/ 2 ) 3 


23. (z 4 ) 3 

24. (w 6 ) 3 

25. (ar 5 ) 6 

26. (i/ 3 ) 7 

27. (m 4 ) 9 

28. (— a 2 b) 2 

29. (- 2 abf 

30. (2xyf 

31. (3 xy 2 zf 

32. (4 ra?i 2 ) 3 

33. (— 2 a^?/) 3 


34. (— 4 a 3 6) 3 

35. (- Ja) 3 

36. (-fed 2 ) 3 

37. (fa*/ 2 ) 3 

38. {x 2 y 2 w) 4 

39. (— a 2 b 3 ) 5 

40. (a 3 ) n 

41. (6 m ) 4 

42. (zV) n 

43. (rV) n 

44. (x m y n y 






334 ALGEBRA 


45 . 

(?)’ 

49 . 

0 

63 . 

(1)' 

-■ (?)' 

46 . 

(t)‘ 

50 . 

/ »l\° 

w) 

64 . 

& 

•* (?)' 

47 . 

/3 c 2 \ 3 
\5d ) 

61 . 

<$)' 

65 . 

(ft 

“ ©■ 

48 . 

/ 2 m 3 \ 3 

\ 3 n 2 ) 

62 . 


56 . 

/x m \ 2 

\y n ) 

»■ $r 


218. Cube and other roots appear in pure and applied mathe¬ 
matics. 

Just as a number has two square roots , so it has three cube 
roots , four fourth roots , and n nth roots. 

Thus, consider the fourth roots of 16. 

1. + 2 is one, since (+ 2) 4 = 16. 

2. — 2 is one, since (— 2) 4 = 16. 

3. +2 i is one, since (+2 i) 4 = 16 i* = 16 X 1 = 16. (§ 189) 

4. — 2 i is one, since (— 2 i) A = 16 i* = 16 X 1 = 16. 

Of these four roots, + 2 is called the principal fourth root of 16. 

The principal even root of a positive number is a positive 
number. 

I 

One of the three cube roots of + 8 is + 2, since (+ 2) 3 = 8. 

The other two are complex numbers. + 2 is called the principal 
cube root of + 8. 

One of the three cube roots of — 8 is — 2, since (— 2) 3 = — 8. 

The other two are complex numbers. — 2 is called the principal 
cube root of — 8. 

The principal odd root of a positive number is a positive 
number; the principal odd root of a negative number is a 
negative number. 

The principal nth root of a number x is indicated by the 
symbol ~Vx. The number n is called the index of the root. 
The number under the radical sign is called the radicand. 


LITERAL AND NEGATIVE EXPONENTS 335 


We know that Vx 6 = x 3 , since (x 3 ) 2 = x 6 ; 

that v' - y* = — y 2 , since (— y 2 ) 3 = — 2/ 6 ; 

that ^81 x 12 y 9 = "V / 3 4 x 12 ?/ 8 = 3 x 3 ?/ 2 , since (3 x 3 ^ 2 ) 4 = 81 x 12 i/ 8 ; 

that v' — 32 xV° = v^(— 2) 5 xV° = — 2 xy 2 , 

since (—2 xy 2 ) 5 = — 32 xV°- 

Rule. — To find the principal root of a monomial: 

1. Write each factor of the monomial with an exponent 
which is its original exponent divided by the index of the root. 

2. If the radicand is positive, the root is real and positive, 
whether the index is even or odd. 

3. If the radicand is negative, the root is : 

a. real and negative, if the index is odd; 

b. imaginary, if the index is even. 

(In this chapter, no imaginary roots occur.) 

Historical Note. — A symbol for extracting a root did not appear 
until the fifteenth century. In Italian mathematics, the first letter of 
the word Radix, meaning the root, was used to indicate the square 
root: thus, R. Presently there were used R.2“, R.3 a , etc., to indicate 
the square, cube, and other roots. Chuquet, a French mathematician 
of about 1500, used R 2 , R 3 , etc. 

In Germany, a point was placed before a number to indicate that its 
square root was to be taken. Two points were used to indicate the 
fourth root, and three the third root. Reise, 1492—1559, replaced the 
point by the symbol V> to indicate the square root, and Rudolph, 
1515, used the symbol, VV, for the fourth root. Stevin, 1548-1620, 
used the better symbols: V(D> V(D> etc. Girard, 1590-1632, used \/, 
\/, etc. Descartes used the vinculum to indicate what numbers were 
affected by the root. 

EXERCISE 164 


Give the principal root indicated below: 

1. 6. V- 27mV 

2 . ^27 6 . y /64 /‘if 

3. 8 a 6 7. 

4. </- 125 r 12 8. </ - 2 -V rV 


9. 

10. "vlO a 4 6 4 

11. v^l X 4 j/ 8 

12. ^256 ra 4 
















ALGEBRA 


336 


13. 


17. 

V- ■sV 1 ' 5 * 5 

21. 

^ “ izt 

14. 

03 

to 

© 

18. 


22. 

x 4m y 8n 

15. 

V - 243 aV 

19. 

V64 a 6 6 12 

23. 

</ — x bn y 10 

16. 

- 32 zV 5 

20. 


24. 

</ x 12r y 6r 

25. 

v 

30. 

Xy 12 

35. 

ja 

V» 

26. 

1 8 jl' 

\ J12 

31. 

*/ m 10 

^ 7l 15 

36. 

3 /^ 

Xtfm 

27. 

3 / 64 m 6 

32. 

a/? 

37. 

®/- z 5r 


* 27 » 3 

\y6 


' 2/ 10 * 

28. 

€ 

33. 

4 /l6 tt 4 
\ 6 8 

38. 

ijx ia 

V 

29. 

<1 f 

34. 

«/— 32 m 5 

^ 7l 10 

39. 

n jx mn 


219. Negative numbers, fractions, and zero also are used 
as exponents. So far, however, such expressions as a -3 , a^, 
and a 0 have not been given any meaning in this text. 

Observe that a~ z cannot mean that a is used as a factor - 3 times, 
since doing anything - 3 times does not have meaning. 

220. Meaning of a fractional exponent. 

If a 5 is to obey the law of multiplication on page 332, then 
a* • a f • a? = a 5 . 

That is (a*) 3 = a 5 , or cfi = v'a 3 . 

This suggests the 

Definition. — In a fractional exponent, the denominator indi¬ 
cates a root of the base, and the numerator a power of the base. 

Thus, a* = . a j so 
























LITERAL AND NEGATIVE EXPONENTS 337 


EXERCISE 165 

Express with radical signs and find the value of: 


1. 8* 

9. 125* 

17. (^-a 6 )* 

2. 16* 

10. (- 1000)* 

18. (-27 m 3 )* 

3. 32* 

11. (- 32 a 5 )* 

19. (- 243 a 5 )* 

4. 64* 

12. (- xV)* 

20. (- 125 x 6 )* 

5. 64* 

13. (- 64 a 3 ft 6 )* 

21. (*TO 4 )* 

6. 100* 

14. (81 TO 4 )* 

22. (- T V« 3 )* 

7. 81* 

16. (- z 10 ® 5 )* 

23. (i-Jt to 6 )* 

8. 256* 

16. (to 12 )* 

24. (-*x 10 )* 

Express with radical signs : 


25. 3* 

29. (-3 x)* 33. 5 • a* 

37. x* 

26. 4* 

30. (3 to)* 34. 2 aft* 

38. (- y)* 

to 

^1 

CN 

31. (-3 to)* 35. (2 aft)* 39 _ y i 

28. (: 2x )* 

32. (5 a)* 36. 



Express with fractional exponents: 

40. Vm 3 42. \^3 a 2 44. 46. 5>/a 2 48. 2 aV^m 2 

41. </x 2 43. 3 </a 2 45. </tfb 3 47. 3</v 2 49. 3 b 2 Va 


221. Meaning of a zero exponent. 

If a 0 is to obey the multiplication law, then 
a m • a 0 = a m+0 = a m . 
a 0 = a m -f- a m = 1. 

This suggests the 

Definition. — The zero power of any number, except zero, 
is 1. 

Thus, 5° = 1; (5000)° = 1; z° = 1; (- 15)° = 1. 




338 


ALGEBRA 


222. Meaning of a negative exponent. 

If a -3 is to obey the multiplication law, then 
a- 3 • a +3 = a -343 = a 0 = 1. 
This suggests the 

Definition. — a~ n = —• 

a n 


Thus, 


z 4 ’ 




1 _ 1 
( 2) 3 8 * 


EXERCISE 166 


Express with positive exponents and find the value of: 


1. 4 -2 

2 . 3 "* 

3. 2~ 5 

4. 5~ 2 
6 . 8 ° 

6. 6~ 2 

7. (— 3) -2 

8 . (— 2) -3 

9. (— 5)~ 2 

10 . (— 2 )”* 


11 . fi - 3 

12 . 8 ~* 

13. 16“* 

14. 81"* 

15. 2 _1 • 5“ 2 

16. 6° • 

17. 64 • 2“ 5 

18. 9° • 9"* 

19. (— 8)~* 


20. (- 64)~* 

21. (-32)“* 

22. 4"* 

23. 81"* • 6 

24. (- 27)"*-9* 

25. 16"*-64* 

26. 18° • 2“ 6 

27. 3 -2 • 27 

28 . 4r* • 256* 


Rewrite with only positive exponents: 

29. aPb- 4 31. 2T-- 4 33. 2 a“ 2 6 4 35. 10“ 2 • x 5 

30. (3 a:)- 3 32. (2 r)~^ 34. (2 a)~ 2 6 4 36. lO^a^ 3 


223. Negative exponents in fractions. 


Example 1. — = 3 • -i -i- 4 • \ 

4 c z a 6 c 2 


3 


4 _ 3_ # c* 
c 2 a 3 4 


3 c 2 

4 a 3 * 


Observe that a -3 of the numerator becomes a +z in the de¬ 
nominator; that c~ 2 of the denominator becomes c +2 in the 
numerator. 



LITERAL AND NEGATIVE EXPONENTS 339 


This suggests the 

Rule. — Any factor of one term of a fraction may be trans¬ 
ferred to the other term provided the sign of its exponent is 
changed. 

Example 2. — ^ = 

w~H yt 

Example 3. — ^ a J } = 3 a 2 bc~ l d ~ 3 . 


EXERCISE 167 


Rewrite with only positive exponents : 



■x-*y 

" z 2 

5 0T 3 

y- 2 

5 m 2 n~^ 

’ 6 r 2 

10 . 

k 

IQ-15 

3 m 

_ mn 

0 . -- 

4 a~*b 2 

11 . 

10-V 

n~ 2 

10- 5 

' 5 AV 


f 

4 a~ 2 

6 ** 

„ 8 a*6~* 

12 . 

4 ■ 3 _1 7r 

'■ v 

' (2 y)~ 2 

ci 


r -3 

Write without any denominator: 



if* 

2 mV 

17. 

19. 

5 a 3 6 -2 

y 3 

r 5 

m 4 


2“ 1 c- 2 


16. 

18. —- 

20. 

6 a* 

d~ 2 

2 4 

y* 




224. It can be proved that the five fundamental laws of page 
332 apply for any rational exponent. The laws will be assumed 
in this text. 

EXERCISE 168 

I. The multiplication law. (Page 332.) 

Example. — o 5 • a ~ 2 • a 0 • dfc = a 5 2+0+ ^ = a 3 ^. 

1. Express in words the multiplication law. 



















340 


ALGEBRA 


2. Multiply each of the following numbers : 

r 3 ; 5~ 3 ; rV; r^ -4 ; r m s n 
a. by r 4 6. by s -3 c. by rV d. by r~ 2 s~ 2 

3. Multiply each of the following numbers : 

2/*; aV; * a 2/ 6 

a. by a;* 6. by ?/* c. by a^y* d. by 


4. Multiply each of the following numbers : 

ra - *; 

a. by m 2 6. by c. by d. by ra _1 ?i 2 


Multiply: 

6. a? — a?b* + by a? + 6 ¥ 

6. 2 z -2 — 3 a:" 1 - 4 by 3 ar 1 + 5 

7. a^ + x^y % + %*y^ + 2/* by x^ — y* 

8. x~ 2 + x~ l y~ l + y~ 2 by a: -2 — x~ l y _1 + y~ 2 


9. (a:^ + y*)(%* ~ yty 

10. (x~ 2 + y~ 2 ) (x~ 2 - y~ 2 ) 

11. (a -3 + 4) (a -3 + 4) 

12. (x m - y n ) 2 

13. (a; -3 + 6 y _1 ) x~* + 2 y _1 ) 

14. (a* - 2 6*) (a* + 3 6*) 


15. (a; + z -1 ) 2 

16. (z“ + i/ 6 )(a° - y b ) 

17. (z° - 3 b)(x a + 5 6) 

18. (x m + 4 i/ n ) (a: m - 7 ?/ n ) 

19. (a 1 - 7)(a f + 9) 

20. (c _i + <r*)(cf* - <r*) 


II. The division law. (Page 332.) 

Example. — rri^ -r- _ m 3 a. 

21. Express Law II in words. 


22. Divide each of the following numbers : 

t 7 ; r 8 ; t m ; t*; f* 

a. by J 3 6. by t~ 2 c. by 2* 


d. by t 


LITERAL AND NEGATIVE EXPONENTS 341 


23. Divide each of the following numbers : 

ary; x m y n ; ; x~^y~* 

a. by x^y 3 b. by x~ 2 y~ l 


c. by x*y 




Divide: 

24. a 2 -f 2 a + 1 by a 4 

25. 6 x -2 — 8 or 3 + 4 or 4 by 2 or 1 

26. ra 2 — m + 1 by irfi 27. 3 c 2 — 6 cd — 9 d? by 3 cd 

28. 4 r* — 6 rs — 10 s* by 2 rV 

29. a — b by + b * 31. a 2 — b 2 by a* — b * 

30. a — 6 by — b* 32. a; 2 + y 2 by 

33. c 3 — d 3 by c* — d? 

III. Power of a power law. (Page 332.) 

Example. — = x~% ' % = x~ 2 . 

34. What is the value of each of the following numbers ? 

(r*) n ; (y- 4 )"; (z *) n ; (w“*) n 

when n is: a. 2 6.-3 c. \ d. — J e. — -J 

35. (10 2 ) 3 = ? 37. (10 2 - 6 ) 4 = ? 39. (10 2 - 56 )* = ? 

36. (10 1 - 5 ) 2 = ? 38. (10- 1 - 3 ) 2 = ? 40. (10- 9 - 824 )* = ? 

IV and V. Power of a product and power of a quotient. 

(Page 332.) 

Example 1. — (x~ 3 y%) 4 = x~ l2 y 2 . 

ExamV le2.- (g) = £ = 


41. Express Law IV in words. 

42. Express Law V in words. 

43 . Give the simplified value of each of the following numbers : 


(a 2 6“*) n ; (mr 2 p *) n ; (x *y*) n ; 
when n is : a. 3 b. — 2 


(r a s~ b ) n 


d. 


1 

2 


C. 





342 


ALGEBRA 


Find the vaue of: 
44. (- 125)* 


Solution. — 

(- 125)* = ((- 5) 3 )* 

- (- 5) 2 

= 25. 

46. 9* 

62. 81* 

69. 

(36 a 4 )* 

«i» 

00 

<£> 

63. (- 8)* 

60. 

(- 1000 c 3 d 6 )* 

47. 16* 

64. (- 27)* 

61. 

(- 8)-* 

48. 25* 

66. (9 it 2 )* 

62. 

(16)-* 

49. 27* 

66. (- 27 a 3 )* 

63. 

(25)-* 

60. 64* 

67. (- 32 I 5 )* 

64. 

(-32)-* 

61. (- 125)* 

68. (16 x 8 )* 



66. What is 

the meaning of: 



a. 10- 5 

b. 10 26 c. 

10 1 - 5 

d. 10 1 - 75 

66. Simplify 

10 21 X 10 1 - 4 
' a ' 10 1 - 2 

, 10 1 - 25 X 10 3 - 40 

1Q2.6 

67. Multiply 

each of the following numbers: 



10 125 ; 10 21 ; : 

io 9 - 73 


a. by 10 

b. by 100 


c. by 10* 3 


68. Divide each of the numbers in Example 67: 

a. by 10 b. by 10 105 c. by 100 

69. Find: a. (10 1 - 6243 )* b. (10 3 - 4275 )* 

70. Find: a. (10 mo ) 2 b. (10 2 - 4771 ) 3 




XX. RADICALS 


225. A radical is the principal root of a number, indicated by 
a radical sign: as V^a, or Vx + l. 

If the indicated root can be obtained, the result is a rational 
number; if it cannot be obtained, it is an irrational number. 

226. The order of a radical is determined by the index of 
the radical. (Page 334.) 

Thus, a cube root is a radical of the third order; 

a fourth root is a radical of the fourth order. 

227. Radicals of the second order are most important and 
will be emphasized. Give the approximate decimal value of 
final results for radicals of second order. In many lists, in 
this text, the examples involving radicals of higher order are 
given separately and may be omitted unless the teacher has 
special reasons for teaching them. 

228. Two fundamental principles of radicals : 

T f-v/rV* = x 

Thus, ('V / 19) 3 = 19; (v^) 5 = 3. 

Really this is only the definition of n th root; that is, the 
nth root of x is the number whose n th power is x. 

II. Vab = V~a -Vb. Thus, </T^ = • < / 9. 

Proof. — y/ab= (a&)» = a” ■ b* = Va ■ Vb. 

Expressed in words : the nth root of the product of two or more 
numbers equals the product of the nth roots of the numbers. 

343 



344 


ALGEBRA 


REDUCTION OF A RADICAL TO SIMPLEST FORM 

229. Reducing a radical to an equivalent radical of lower 
order. 

Example 1. — \ / 125 = = (5 3 )^ = 5* = 5^ = V~5. 

^125 = V5 = 2.236+. 

Example 2. — V64 = V2 6 = (2 6 )^ = 2* = v / 2 2 = v^. 

EXERCISE 169 


Reduce to radicals of lower order (see § 227) : 


1. vS 

7. V36 

13. v / 225 

19. </ 12b a?b 3 

2. ^256 

8. V900 

14. \/l44 

20. ^243 w 6 

3. v'8 

9. \/ 243 

16. \/32 

21. V / 25 c 4 

4. \ / 49 

10 . v / 27 

16. \/64 

22. Vl6m 8 y 8 

5. Vlii 

11. V64 

17. Vl25 

23. 27 a 3 // 

6. \ / 25 

12 . 7m 

18. V'E) aV 

24. ^169 zV* 


230. Removing a factor from the radicand. 

Example 1. — V75 = V25 • 3 = V25 • Vs = 5V3. 

V75 ^ 5V3 =T5(1.732) = 8.66j) 

Example 2. — v 7 96 a 5 6 12 c 8 = ^(32 a 5 6 10 c 5 ) (3 6*c®) 

= ^32 a b b l0 c 5 • v/jfW 
= 2 ab 2 c • ^TPc 3 . 

Rule. — To simplify a radical by removing factors from the 
radicand: 

1. Separate the radicand into factors as many of which as 
possible are perfect powers corresponding to the indicated root. 

Thus, when it is a square root , separate the radicand into as many 
perfect square factors as possible; when it is a cube root, separate the 
radicand into as many perfect cube factors as possible; etc. 

2. Find the indicated root of each of the perfect powers. 
Obtain the product of these roots. 














RADICALS 


345 

3. Multiply the result of Step 2 by the indicated root of the 
remaining factors of the radicand. 


Example 3. — Vl28 x*y 8 = V2 7 • y 7 • x?y = 2 yVx?y. 

EXERCISE 170 

Simplify by removing factors from the radicand (see § 227): 


1. 

Vis 

6. 

V75 

11. 

V45 m 4 n 3 

16. 

V 98 rs 3 

2. 

V27 

7. 

v 7 63 m 2 

12. 

V32 Sy b 

17. 

V 96 m 2 n 4 

3. 

V200 

8. 

V28 a 2 b 

13. 

V99 ab b c 3 

18. 

Vl08 a 7 

4. 

\/l25 

9. 

Vl2 A / 2 

14. 

V m 3 n 2 

19. 

V300 m 5 n 2 

6. 

V48 

10. 

V 80 7s 

16. 


20. 

V363 x 6 

21. 


24. 

-v / 32m 4 

27. 

^32 a: 5 ?/ 6 

30. 

V 16 x 3 m 2 

22. 

^54 y b 

26. 

v 7 162 a 2 6 4 

28. 

^ 64 a; 7 ?/ 

31. 

\/72ab i 

23. 

v 7 128 t e 

26. 

^80 <?& 

29. 

4^250 a 3 6 4 

32. 

v / 243 x 2 m 

33. 

/60 a 2 

36. 

3 / 4 a 3 

37. 

3 / 8 xy 

39. 

5 / 3 ^ 

^9 y 4 

^27 *V 

' 125 a 6 m 3 

\ a 10 y 5 

34. 

199 mn 2 

36. 

4 / 2a 

38. 

4/5 X 4 J/ S 

40. 

s/64 j/ 5 

^ 16 p 4 

•si xy 

V 256 

^ X 10 


231. Changing a fractional to an integral radicand. 

Example 1. - JX = J 5 ’ 3a = = J-VUZ. 

’12 a ’12a* 3 a ’36 a 2 6 a 

Observe that 12 a is multiplied by 3 a to change it into 36 a 2 , the 
“ smallest ” perfect square which is a multiple of 12 a. Then the 
numerator is also multiplied by 3 a to avoid changing the value of the 
fraction. 

Rule. — To change a fractional to an integral radicand : 

1. Multiply both terms of the fraction by the number which 
will make the denominator the least perfect power of degree 
corresponding to the order of the radical. 

2. Simplify the resulting radical as in § 230. 













































346 


ALGEBRA 


Note 1. — This process is called rationalizing the denominator. 

Note 2. — If it is a cube root, the denominator must be made a perfect 
cube; if it is a fourth root, it must be made a perfect fourth power; etc. 


Example 2. 

II 

"1 

II 

** 

1 3 3 -: 
1 2 3 

*•.*- Va*. 

•a 6 2 a 2 




EXERCISE 171 




Express with integral radicands (see § 227) : 



1. 

4 V 

*-4 

'■ 4h 

10. 

J- I* 

*lln 

13. 

Jm. 

*200 

2. 

4 

‘4i 

•■4i 

11. 

(5Z 

'8 s 3 

14. ' 

J 13 

’50 cd 

3. 

4 

*24 


12. 

*20 t 

16. < 

C* 

^ S 

^ 00 

16. 

4 

19.^ 

*16 


26. 

* a 

28. - 

6/ ay 
*32z 6 

17. 

4 

20. xfe- 1 

*25 

23. \j—. 

*8 n? 

26. 

5 / 5 ^ 

* 16 

29. - 

*x-y 

18. 


21. 

*2 X 5 

24. 

’27 n 3 

27. 

12ri* 
* 9 a 9 

30. ^ 

yl c + d 


232. To reduce a radical to its simplest form: 

a. Reduce it to an equivalent radical of as low order as 
possible, as in § 229; 

b. Remove from the radicand any perfect power factors of 
the radicand, as in § 230; 

c. Change the radicand to an integral radicand, as in § 231. 

233. Addition and subtraction of radicals. 

The sum of two radicals like V2 and Vs can be indicated; 
thus V2 + V 3 . Or, each can be expressed decimally and the 
approximate sum be obtained; thus, 

V 2 + V3 = 1.414 + 1.732 = 3.146. 
















RADICALS 


347 


Similarly the sum of a/ 2 and a/2 is V2 -j- v/ 2- 

In order that radicals may be combined, they must be similar 
radicals. 

Similar radicals are radicals of the same order, which have 
the same radican^. 

Thus, 2 VWa and 3Vl3a are similar radicals. They differ only 
in their coefficients. 

Also, VS and Vax 2 are similar radicals; 

but yftfx and Vax 2 are not similar radicals. 


Rule. — Xo add or subtract radicals : 

1. Change them to equivalent radicals of simplest form. 


(§ 232) 

2. Combine similar radicals; indicate the addition of dis¬ 
similar radicals. 


Example . — ^-^ 24 +^ 54 =^“ ^ 

= \</2 - 2v / 3 +3V2 

= 3i^2 - 2 v / 3. 


EXERCISE 172 

Simplify by combining like radicals: 


1 . V32 + >/72 

2 . 2V\Sx - VdOx 

3. a/ 128 a - 3 V8a 

4. V7 + iVl8 

5 . — V^m ^ 

6 . Vf? + 

7. 3 aA/l6a - V144"5 


10 . ^16 + v / 250 

11 . v'M z — v' 16 i 

12 . V 192 y 2 - V / 24/ 

13. v'f + v"! 

14. 

15. 3 ^ + 5a/3"? 

16. + 'V 7 256 a® 


8 . 2 WWrf - 3 yVl6? 17.^729 -^3 

9. «V98m 3 n - mVZ2mn 3 18. “^2 + Vl28 













348 


ALGEBRA 


19. 

20 . 


V: 


2l _1 f a ~h 1 _ 

a 4“ 1 * a — 1 


fjr + V 2 xV 
x b y b 



Va 2 - 1 


21. K \/x 5 + Va? 


MULTIPLICATION AND DIVISION OF RADICALS 

234. Changing radicals of different orders to equivalent 
radicals of the same order is necessary before radicals can be 
multiplied or divided. 

Example. Change V2, V3, and V 7 ^ t° radicals of the same 
order; and determine which is the greatest number. 

Solution. — 1. V2 = 2^; = 3!; ^5 = 5} 

2. Change the fractional exponents to their L. C. D. 

2^=2^; 3^ = 3 A; 5* = 5*. 

3 2* = 2* = 'VW = 1 V64 ] 

3^ = 3^7 = 1 Vs i = 1 V81 1 is the largest number. 

5* = 5* = ‘VP = 1 Vl25 J 

Rule. To change radicals of several orders to equivalent 
radicals of the same order: 

1. Express the radicals by means of fractional exponents. 

2. Change the exponents to their lowest common denominator. 

3. Rewrite the resulting expressions with radical signs. 


EXERCISE 173 

Change the radicals in each example to radicals of the same 
order; select the greatest in Examples 1, 2, 7-12. 


1. and V 2 

2. 's/\ and 

3. VTa and 6 a? 

4. v^ll b and ‘C 7 124 b 

5. V8 a: 3 and X^7 x 4 5 6 

6. 's/a 2 }) 2 and \/ab 


7. V5 and "V 7 124 

8. V 7 ^ and V % 

9. </ 2 , and 

10 . </ 6 , </ 2 , and V 3 

11 . V2, ^5, and ^3 

12. V2, v 7 !, and v 7 8 










RADICALS 


349 


13, 's/c + d and Vc + d 

14. \/x — y and Vx + y 


15. a — 2 and ^a + 2 


16. \/m + n and 'Vm — n 

235. Multiplication of radicals. In § 224, we have assumed 

that ( ab )* = a» • bn, or a* • b* = (a6)». _ 

Written with radical signs, this becomes Va • V6 = Va6. 
That is, the product of the nth roots of two numbers equals 
the nth root of the product of the two numbers. 

Thus, 2V3 • 3V6 = 6V3~6 = 6VF“2 = 6 • 3 V 2 = I 8 V 2 . 

4^2 • 5 ^ = 20 ^ = 20 • 2 = 40 . 

V2-Ay2 = 2i-2i = 2*-2*=v / 2 5 -' / 2 5 = V&. 

Rule. — To multiply radicals : 

1. Change the radicals, if necessary, to radicals of the same 
order. (§ 234) 

2. Place the product of the radicands obtained in Step 1 under 
the common root. 

3. Reduce the result of Step 2 to simplest form. 


EXERCISE 174 


Find the following products (see § 227) : 


2V / 6~a -5V3a 

1. V3 • 

Vl8 

7 . 3V7 • 5V7 

13. 

2 . V2 • 

Vl4 

8 . 2Vl2 • 3V3 

14. 

2\ // 6~x • 3V&X 

3. 2V7 

• V21 

9. (V5) 2 

15. 

(V^a -f- b) 2 

4. 3V6 

■2Vl2 

10 . ( 2 V 3) 2 

16. 

(Vx - 3) 2 

6. 3V3 

■V27 

11. (4V5) 2 

17. 

(3 Vy + 2) 2 

6. 5V20 • V^SO 

12. (6V2) 2 

18. 

(4V^5) 2 

19. Multiply 2V3 + 3V2 by 3V3 - V2. 



Solution. - 2V3 + 3V2 

3V3 _ v2 

18 + 9V6 
- 2V6 - 6 

18 + 7V6 - 6 = 12 + 7V6 

/. (2>/3 + 3V2)(3>/3 - V2)' = 12 + 7(2.44) = 12 + 17.08 = 29.08. 














350 


ALGEBRA 


> 


Find similarly the following products : 


20. (2 -VS) (2 + VS) 

^ 21. (7 - V§)(6 + VS) 

22. (VS + 4)(V5 - 3) 

23. (2VS + 7)(iVS - 7) 

24. (VS - 4) 2 
26. (2V2 - 7) 2 

26. (2V2 + 3) 2 

27. ( 2 V 2 + 3)(4v / 2 - 5) 

36. Find WS • VIES. 


28. (3v^ — 5)(sVS + 5) 

29. (VSS — Vx)(VSH + Vx) 

30. (VIS - VS)(VIS + VS) 

31. VI + 1 • VI + 1 

32. (Va - 1 + l) 2 

33. (VI - 5 - 3) 2 

34. (Vy - Vy + 3) 2 

36. (Vx + 3 + Vx - 5) 2 


Solution. — 1. v / 9 ~a 2 • Vl 5 a = (3 2 • a 2 )I .(3.5 ,,)t 
2- = (3 2 • a 2 )» • (3 • 5 • a)* 

3. = -VZ' ■ a A ■ 3 3 - 5 3 ■ a 3 . 

(Complete the solution.) 

Find: 


37. ^9 • 

38. 3^16 • ^36 

39. </%xy • </21 x 2 y 3 

40. v / 54 x 2 • v^9 xy 

41. 3vT^ . 2v / 40 a 3 

42. 3^27 x b • 81 a: 2 

V 43. V2a-\/T^ 

44 . VS • y/~x? 


45. \Z~2 m 2 • V^8 m 3 

46. VlO • ^25 

47. ^16 rt 2 • VTf 

48. 2 aV^ • 5\^4~x 

49. x 2 y 2 • \^2 x 4 y 3 

50. 

61. ^ 9 a- 2 ?/ • V15 a:?/ 3 

52. a/M-V® 


236. Division of monomial radicals. (See § 227.) 

Consider VS + VS. It would be possible to obtain the 
approximate value of the result by using the approximate 
decimal values of V5 and V2. 


V6 = 2.449 
V2 1-414 


1.7319. 


Thus, 



















RADICALS 


351 


This solution requires the computation of two square roots, and the 
division of one decimal by another. 

A preferable solution follows : 

= V 3 = 1.732. 

V2 ' l 2 

This solution requires only one square root, the division of a decimal 
by an integer, and is more accurate. 


Similarly, 


_v^8_ 

^18 


3 1 - 

'9 


^ 12 . 


Rule. — To divide one monomial radical by another: 

1. Change the radicals, if necessary, to equivalent radicals 
of the same order. 

2. Write the quotient of the radicands obtained in Step 2 
under the common root. 

3. Simplify the radical resulting in Step 2, by the methods 
summarized in § 232. 

EXERCISE 175 


Perform the indicated divisions (see § 227): 


1. V27 4- V9 

2. V21 4- V7 

3. 9V20 -5- 3V5 

4. V50~x -5- V2x 

5. 3 aV 90 4- oV3 

6. V3 - 4 - V| 

7. 20VicV 5V^ 

8. 14V30 a? v7V6a 

9 . 4 xV 24 x 3 m 4- 8 xVs x 
10. Vl4 y 3 4- Vjr y 2 


11 . 121 Vcd 4- \lV\d 

12. V2 4- V3 

13. V5 4- V(3 

14 . V87 4 - a/77 

15. V33 m 2 n 4- Vl8 n 

16. (6Vl5 - 8VT2) 4 - 2A/3 

17. (5Va* + 20Vx) 4- 5Vx 

18. (Vg + 2Vl2) 4- V3 

19. (4VI2 - 3V18) 4- VG 

20. (5V20 - Vl5) 4- V5 












352 


ALGEBRA 


21. V / 250 -H \/2 

22. ^84 -r- ^21 

23. V / 39~? V 7 13 ic 6 

24. V 7 48 z 8 ^3a? 

25. \^125 a 2 6 2 -f- V 7 25 6 2 c 
31. Find V3 -i- \^9 


26. 4 ^^12 a: 2 ?/ 4- ar^ 7 6 xy 3 

27. 3 mV 7 12 s 3 -f- 12 mV 7 i / 

28. 5 c 2 V / 6 cd -5- X V^^ c 2 d 

29. V8 a: 3 2/ 2 V4*n/ 2 

30. -Vax 2 y z -f- Vaxy 2 


Solution. — 


V3 = _3f_ = _3$_ = ^33 = ? 
^9 (3 2 )? (3 2 )* ^3* 

(Complete the solution.) 


32. 3 -f- 'V / 3 

33. 2 -r- </2 

34. 12 -r- </Z 

35. 6 -i- ^36 

36. x -5- yx 

37. V3 4- \^3 

38. V2 + X^2 


39. ^4 -r- 

40. Vl6~a -b \/32~a 

41. ^18 re 2 V6^ 

42. </J^ - V 7 ! _ 

43. ^9 x 2 y ^81 xy 

44. 10v 7 216 -f- 2 a/6 

45. 3v / 90 or 2 ?/ -r- 2V 7 6 xy 2 


237. Division by a binomial quadratic surd. 

Consider -——. The approximate result can be found in 

2 + V3 

decimal form as follows : 


2 + V3 2 + 1.732 3.732 

This solution requires the division of an integer by a decimal. 

A preferable solution follows : 

We observe that 2 + V3 multiplied by2-v / 3 = 4_3=i < 
Multiply both terms of the fraction by (2 — V3) : 

then _1— = _ ~ = 2 - V3 = 2 - V§ 

2 + V3 (2+V3)(2-V3) 4-1 1 

••• - -—— = 2 - 1.732+ = .267+ 

2 + V3 





















RADICALS 


353 


In this solution, division by a decimal is avoided. The de¬ 
nominator is multiplied by a number such that the product is a 
rational number. For this reason, this process is called ra¬ 
tionalizing the denominator of the fraction. 

If the denominator is of the form a -h by/c, it is multiplied 
! by a — bVc. These two surd expressions are called_conjugate 
surds. Observe ( a + by/ c) {a — by/c) = a 2 — b 2 y/ c 2 . 

(a + b Vc) (a — b y/c) = a 2 — b 2 c. 

Their product is a rational number. 

Rule. — To divide a number by a binomial quadratic surd, 
multiply dividend and divisor by the conjugate of the divisor, 


and simplify the result. 

3\/2 4 . l 

Example. — Rationalize the denominator of —^= --• 

3V2 + 1 = (3V2 + 1)(2V2 + 1) = 12 + 5V2 + 1 
2 V 2 - 1 ( 2 V 2 - 1 )( 2 V 2 + 1 ) 8-1 

. 3V2 + 1 = 13 + 5V2 = 13 + 5(1.414+) = 2 86< 

2 V 2 - 1 7 7 


EXERCISE 176 

Perform the following divisions by rationalizing the divisor: 


10 

V . 

6 

11. 

a/10 

3 + V 7 

6. 

3 + 

2 V 5 


a/5 - V2 

3 

7 

5 + 

V3 

12. 

Vx — 3 
- - /- 

V8 + 4 


2 - 

v§ 

y/x + 3 

4 

Q 

2 + 

V5 

13. 

yfa + y/b 

9 - V2 

O. 

4 - 

V2 


Va - a /b 

3 

t 9. 

V6 

-V3 

14. 

2VI - 5 

5 - 2 V 2 


V6 

+ V 3 


V§ + 5 

5 

i A 

V3 

+ V 5 

15 

Ve + 2 V 3 

CO 

1 

CM 

\ 10. 

Vs 

-Vs 


2^6 - V3 

























354 


ALGEBRA 


16. 


2V2 + 5 

VS -2 


V a — 1 -J- 2 

V a - 1 + 1 


3 + V2 lg c - 

3 — V~2 c + aVb 


Vx + y — Vx — y 
Vx+y + Vx-y 


238. Involution and evolution of monomial surds. 

Example 1. — (^12) 3 = (12*) 3 = 12* = 12* = Vl2 = 2Vs 
( ^12) 3 = 2V§ = 2(1.732+) = 3.464+ 

Example 2. — V</27 3 ? = {(3 3 • s 3 ) *}* = (3 3 • x 3 )* = 3*a£ 

V = Vsx 


EXERCISE 177 


Simplify the following expressions : 


1 . (^7)* 

8 . vWs 

15. \ V 243 a 10 

2. (Vv? 

9. VVa*b 3 c 6 

16. V^9a 16 

3. (Vl6) 3 

10. Vv'T? 

17. Vv / 27a 3 6 6 

4. (Vl2) 4 

11. i/VSQm 6 

18. (VaW) 4 

6. (^m) 4 

12. VV128 

19. (2^)’ 

6. (2x< / 7 2 )* 

13. -\/VSQ 

20. (3 ra 2 V / »?) 3 

7. (y-VZxf 

14. i/Vl25 

21. vVl6 x 4 y 2 


22. V Va 2 - 8 a + 16 24. (VSx - 2 ) 3 


23. a/^ 4 a: 2 - 4 ar?/ + y 2 26. (V(2 r - 5 s) 2 ) 2 

239. The jquare roots of a binomial surd expression of the 
form a + Vb can often be found by inspection. 

Preparation. — Observe that (V2 — Vs ) 2 = 2 — 2V^6 + 3 
= 5 - 2V6. 

Notice that the square of a binomial surd is a binomial surd; that 
5 is the sum of the two radicands 2 and 3, and that 6 is the product 
of the two radicands 2 and 3. This suggests the 




































RADICALS 


355 


Rule. — To find the square root of a binomial surd expression 
of the form a + Vb : 

1. Change the surd part so that its coefficient is 2. 

2. If possible, separate the rational term into two numbers 
whose product shall be the radicand obtained in Step 1. 

3. If Step 2 has been possible, extract the principal square 
root of each of the numbers obtained in Step 2, and connect 
them by the sign of the surd term. 

Example. — Find the square root of 22 — 3V32. 

Solution. — 1. 22 — 3V32 = 22 — V9 • 8 • 4 (See Note below.) 


2 . 

3. 

4. 

5. 

6 . 

7. 

Check. — 


22 - 3V32 = 22 - 2V72 
22 = 18 + 4; 72 = 18-4 


V 22 - 3V32 = V 22 — 2 V72 


= Vi8 + 4 - 2V72 

= Vl8 - Vi 
= 3V2 - 2 

(3V2 - 2) 2 = 18 - 12V2 + 4 

= 22 - 3 • 4V2, or 22 - 3V32. 

Note 1. — Observe the change from 3V32_to V 9 • 8 • 4. 3 before 
the radical is the same as V9; and V9 • V32 = V9 -8-4. When 
a coefficient of a square r oot i s placed un der th e radical, it must be 
squared. Thus, aV5 = V5 a 2 ; QVx = V36x 2 . 

Note 2.— [—(3V2 — 2)] or [2 — 3V2] also is a square root of 
22 - 2V72. 


EXERCISE 178 


Find the square root of: 

1. 5 + 2-s/6 4. 8 - V48 

2 . 8 - 2Vl5 5. 14 + V132 

3. 12 + 2V35 6. 11 - 4V6 


7. 10 - V§4 

8. 13 + 4VIo 

9. 12 - 8 V 2 


IMAGINARY NUMBERS 

240. An introduction to imaginary numbers was given in 
189, page 285. Review this section. 










356 


ALGEBRA 


241. Powers of the imaginary unit i. 

Since i = — 1, then 


= - 1 


i° = - 


i* = i 2 • i 2 = ( 


i) • (- i) ; 4 = + i 


Therefore the first four powers of i are i, — 1, — i, and 1. 
These four values recur in regular order; thus, i 5 = i; i 6 
= — 1; i 7 = — i; and i s = 1. (Prove these statements.) 

242. Multiplication of imaginary numbers and complex 
numbers. (See § 190.) 

Example 1. — V— 2 • V— 3 = iV 2 • iVs — i 2 \Z& = — 6 

Example 2 . — Find (2 — V — 3) (5 + V— 3). 

SoZwto. — 1. (2 - V^3)(5 + v^3) = (2 - iV3)(5 + iV3) 

2. =10-3 iV 3 - i 2 • 3 

3. = 10 - diVs _ (_ i) . 3 

4. =13-3 iV3. 

EXERCISE 179 

Find the following products : 

1 . V~^~4 • V- 16 
~2 


2. 

3. V^~9 

4. V^5 


V- 18 


8. xV - y • xV - y 

9. aV — b • cV — d 


V- 10 


6 . V- 25 z 2 -V— 4 a: 2 

6. 3V^~5 • 3V^5 

7. 4V^3 • 6V^3 

15. (- 2 + V^2)(-2 - V^2) 

16. (6 - V^3) 2 18. \i(- 1 + V^2)S 2 


10. V — xy • (— V— ab) 

11. (3 + V^2)(3 - V^~2) 

12 . (2 - \/^ 6)(2 + /= 6 ) 

13. (2 + V^5)(2 + 3V^5) 

14. (8 - V^~7)(10 + y/~—7) 


17. (a + V^T) 5 


20 . \U 


19- j*(-l - V^3)^ 
1 + V- 3)1* 










RADICALS 


357 


243. Division of imaginary numbers can be accomplished in 
two ways. 

iVl2 _ _ 2, 


Example 1. — 


V- 12 


iV \3 


In this case, the division is performed directly. 


Example 2. — 


10 


10 

W 2 


10 • i • V2 
i • V2 • i • V2 
10iV2 = 5iV2 

r 2 • 2 - 1 


- 5iVz. 


In this solution, the denominator has been rationalized by multiply¬ 
ing it by itself. 

2 

Example 3. — Perform the indicated division in 


1 + V"^3 


by rationalizing the denominator. 

2 


Solution. — 1. 


2 . 


1 + v- 3 
2(1 - W 3 ) 


3. 


(1 +tV3)(l -W 3) 

= 2(1 - W 3 ) = 1 ~ W 3 . 
1+3 2 


1 +iV3 j 

_ 2(1 - fV3) 
1 - 3i 2 


Observe that the denominator was rationalized by multiplying it 
by its conjugate. Similarly, to rationalize 3 — iVB, multiply it by 
3 + 1 V 5 ; the product will be 9 — i 2 V25 = 9 + 5 or 14. 

EXERCISE 180 

Find the following quotients, or rationalize the denominators : 
V- 42 


V- 36 


W - 48 
2\^"3 


4. 


6. 




V — 40 x 




7. 


8 . 


V- 32 

8 


6\/— 39 


xy 


3V3^ 


V 10 

10 . 

6 


V^3 

VI5 

11 . 

14 

vCTs 

V^2 

V28 

12 . 

24 


V^8 



































ALGEBRA 


358 


3 

5 

10 ^3 + V-5 

1 + V~^3 

16 ' i+V^2 

19 - V3 - 

13 

1 - V'TT 

20 5 + v ^ 

3 - V— ~2 

7 - 1+2V3I 

2 - V- 3 

4 

3-V=2* 

4 + 1-/T2 

2 - V~^2 

I-* 

pc 

CO 

1 

to 

< 

1 

’ 5 - 2V - 2 


244. Application of radicals. In Chapter XIV, irratioi^l 
and complex roots of quadratic equations were found. Check¬ 
ing by substitution in such cases was not recommended at that 
time. It is possible now to check such roots by substitution. 


Example. — Solve and check the equation x 2 + x - 1 = 0. 


Solution. — 1. By the formula, x 

. _ - 1 + y/5. 

2. .. n- - -, 


- - 1 =«= Vl 4- 4 _ - 1 =»= V5 


and 7*2 = 


1 - VE- 


Check. — Substitute 


- 1 + V5 

2 


in x 2 + x — 1 = 0. 


Does 

Does 

Does 


1 - 2V5 + 5 , - 1 + V5 _ ! = 0 ? 
4 2 

6-2V5—2 + 2V5—4 = q? Yes 


Similarly, 


--— will check. 

2 


EXERCISE 181 

Solve and check by substitution : 

1. a 2 — 2x — 1=0 4. x 2 -2x + 2= 0 

2. x 2 + 2x-2=0 6. 2 x 2 -2 x-\-l=0 

3. z 2 - 3 z + 1 = 0 6. ^ + £+ 2= 0 























RADICALS 


359 


7. In trigonometry, certain six numbers occur, of which five 
bear the following relations to the sixth. Calling the numbers 
5 , c, t, S, C, T , 


a. c = Vl — s 2 


b. t = 


Vl 


c. C = - 

s 

d. S = 


e. T = 


Vi - 


Vi - 


If s = find c, t, C, S, and T. 

8. In Example 7, find the values of c, t, S, C, and T when 


IRRATIONAL EQUATIONS 


245. An irrational equation is an equation in which the vari¬ 
able appears under a radical sign or with a fractional exponent. 

The radical sign or the fractional exponent always indicates 
the principal root in an irrational equation. The sign before 
the root indicates the operation to be performed on it. 

Thus, — V x — 5 means “ subtract the principal square root of 
x - 5.” 

Example 1. — Solve the equation x — 1— Vx 2 — 5=0. 

Solution. — 1. x — 1 — Vx 2 — 5 = 0 _ 

2. /. x — 1 = Vx 2 — 5 

3. Squaring both members, x 2 — 2 x + 1 = x 2 — 5 

4. .\ — 2 x = — 6, or x = 3. 

Check. — Does 3 - 1 - V9 - 5 = 0? Does 2 - Vl = 0? Yes. 

Note. — When a single radical occurs in an equation, transpose 
the radical to one side and all other terms to the other side. Then, 
if the radical is a square root , square both members of the equation; if it 
is a cube root, cube both members; etc. 

Example 2. — Solve the equation x — 1+Vz 2 — 5=0. 

Solution. — 1. x — 1 + V c 2 — 5 = 0 

2. + V x 2 — 5 = 1 — x 

3. Squaring, x 2 — 5=1— 2 x + x 2 

4 . 2 x = 6, or x = 3. 

Check. — Does 3 - 1 + ^9-5 = 0? Does 2 + VI = 0? No. 

Therefore 3 is not a root of the given equation. 


















360 


ALGEBRA 


Recall that a number is sought, in solving an equation, which will 
satisfy the equation. The equation expresses certain relations which 
the number must satisfy. Sometimes the relations imposed are im¬ 
possible, and in that case no number can be found which will satisfy the 
equation. Such is the case in this example. In such cases, the root 
is only apparently a root, and the equation is only apparently an 
equation. 

What is the explanation of the solution x = 3? If the original 
equation is compared with the equation of Example 1, it is noticed 
that the only difference is in the sign of the radical; also that in Step 3, 
after squaring both members in both examples, the resulting equation 
is the same. In each example, if the equation of Step 1 has a root, that 
number is a root of the equation of Step 3; but, since the equation of 
Step 3 is the same in each solution, it cannot be asserted in advance 
whether its root or roots are roots of the equation of Example 1 or of 
Example 2. When finally the solution x = 3 is obtained, the question 
arises, is 3 a root of the equation in Example 1 or in Example 2 ? The 
root x = 3 satisfies the equation of Example 1; it does not satisfy the 
equation of Example 2. It is customary to say that, in Example 2, 
the extraneous root 3 is introduced by the method of solution. 

This example makes clear the necessity of checking the solutions of 
equations. 

EXERCISE 182 

Solve and check the following equations: 

1. V7 a -j- 2 -4=0 4. Vl3 -4x = 2 z + 11 

2. 4 - 3 Vy = - 4 5. Vb x 2 - 6 x + 7 = 10 

3. V7x - 3 + 3 = 2 x 6. \^27 x 3 - 27 x 2 + 1 = 3 x 

7. Solve the equation Vx — 2 -f V2 x + 5 = 3. 

Solution. — 1. Vx — 2 + V2x + 5 = 3 

2. V2x + 5 = 3 - Vx - 2 

3. Squaring, 2 x + 5 = 9 — 6 — 2 + x — 2. 

(Complete as in Examples 1-6.) 

Note. — You will get two apparent roots, but one will not check. 
Remember that the radical indicates the principal root. 

8. 'Vy -f- 5 — Vy = 1 


9. Vs - 3 - 1 = Vs - 10 


















RADICALS 


361 


10. V3 m - 11 + V3 m + 10 = 7 

11 . V2/^+ 7 - Vy - 5 = 2 

12. Vz + 15 - Vz - 24 = 13 

13. Vc + 20 - Vc - 1 = 3 14. V4 y - 40 + 14 = 10 


16. 2V3 z — 2 — SVx - 2 = 2 


16. Solve the equation Vy -f 3 — Vy + 8 
Solution. — 1. Squaring both members : 

y + 3 - 2V(j, + 3)(y + 8) + y + 8 = </. 
(Complete the solution as in Examples 1-6.) 
Note. — Observe the term 2V ( y + 3)(y + 8). It is 
2 • Vy + 3 • Vt/ + 8. 

17. Vx + 2 - Vz - 2 = V2~i 

18. V2T+1 = 2 Vi- Vt - 3 


- Vy. 


19. Vz + 5 — V8 — x = V9 — 2 


21 


20. V3 - 2 c - V7 + 6 c = Vl + 2 
. Solve the equation Vz — 6 + Vz = 


Vz — 6 




Solution. — 1. 

z — 6 + Vz[z — 6) = 3. 

(Complete the solution as in Examples 1-6.) 

22 . Vz +3 + ^ 4=5 = 2L 23. Vz + Vz - 9 = 

2 


36 


24. 


Vi 

10 z 


Vi 


Vz — 


VlO z — 9 

V/ + 1 = V< + 3 

vr+3 v< + 6 

26. Vm + 3 + V 3 = 


+ VlO z + 2 = 


VlO z — 9 


25 


27. 


Vj + 2 
Vz 


Vm -f- 3 
Vi 2 


Vz + 2 3 Vz + 2 




























































362 


ALGEBRA 


28. 


9 + Vx = n VSx + 1 - V8x 

9 - 2Vi 5 ' V3x + 1 + V8x 

30. V27x - VlO x + 5 = V20 x + 9 

31. Solve for a;: 3Vaz + a 2 = V 3 ai + 3a 


32. Solve the equation £ = vxl-i 

'g 

a. for l b. for g 

33. Solve the equation V = V2 gi 

a. for g b. for s 

34. Solve for s : — 


-+-<0 


3 _ 

13 


35. Solve for h: L = 7 rrVr 2 + h 2 












XXI. LOGARITHMS 


246. Logarithms are exponents. 

Every positive number can be expressed, exactly or approxi¬ 
mately, as a power of 10. The exponent corresponding to a 
number so expressed is called its logarithm to the base 10. 

Thus, 10 2 = 100; therefore 2 is the logarithm of 100 to the base 10. 
This is written: logiolOO = 2, or more briefly log 100 = 2. 

Similarly, logio35 is read “ logarithm of 35 to the base 10.” 


247. Multiplication, division, involution, and evolution can 
be simplified by the use of logarithms. To make this fact clear, 
the approximate values of some powers of 10 will be computed 
and some examples will be solved. 

1. 10° = 1; 10 1 =.10 ; 10 2 = 100; 10 3 = 1000 1.0000 = lO 0 - 00 

1.7782 = 10°- 25 

2. 10- 5 = 10* = Vl0 = 3.1623 3.1623 = 10°- 50 


lO^J^XlO*- 10X3.1623 =31.623 
10 2 - 5 = 10 1 X10 1 - 5 = 10 X31.623 =316.23 

3. 10- 25 = (10- 5 )* = V3.1623 = 1.7782 
10 1 ' 25 = 10 1 X10- 25 = 10 XI.7782 = 17.782 
1 0 2.25 = 10 i x 10 i.25 = 10 x 17 .782 = 177.82 

4. 10- 75 = (10 15 ) i = V31.623 = 5.6234 
10 1 - 76 = 10 X10- 75 = 10 X 5.6234 = 56.234 
10 2 ' 75 = 10 X10 1 - 75 = 10 X56.234 =562.34 


5.6234 = 10 0 - 75 
10.0000 = 10* 00 
17.7820 = 10 125 
31.6230 = 10 150 
56.2340 = 10 175 
100.0000 = 10 200 
177.8200 = 10 2 - 25 
316.2300 = 10 2 - 50 
562.3400 = 10 275 
1000.0000 = 10 300 


N ote — Examining the table on the right, observe that logio 1000 = 3; 
logio 177.82 = 2.25; logio 5.6234 = -75; etc. 

363 









364 


ALGEBRA 


Example 1. —Find 3.1623 X 17.782. 

Solution. — 1. 3.1623 X 17.782 

2. = 10- 50 X 10 1 - 25 

3. = 10 1 - 75 

4. = 56.234 (in the table) 

5. /. 3.1623 X 17.782 = 56.234 
The solution is approximately correct. 


Check. — 3.1623 

17.782 
63246 
252984 
221361 
221361 
31623 
56.2320186 


Example 2. — Find 562.34 ^ 17.782. 


Solution. — 1. 562.34 -f- 17.782 


2. 

= 10 2 - 75 10 1 - 25 

28 880 

3. 

_ JQ2.75-1.25 

17 782 

4. 

= 10 1 - 5 

11 098 0 

5. 

= 31.623 (in the table) 

10 669 2 

6. 

/. 562.34 17.782 = 31.623 

428 80 


The solution is approximately correct. 


Example 3. — Find (5.6234) 2 X 316.23 177.82. 


Check. — 31.624 

17.7821562.340.000 
533 46 


355 64 


73 160 
71 128 


Solution. — 1. (5.6234) 2 X 316.23 177.82. 

2. = (10- 75 ) 2 X 10 2 - 50 10 2 - 25 

3. = jqi .50+2.50-2.25 = 10 1 - 75 

4. .*. (5.6234) 2 X 316.23 -s- 177.82 = 56.234 
This solution also may be checked by ordinary computation. 


Obviously a more complete list of exponents (logarithms) 
and ability to use them must be of great advantage, for in each 
case the solution by exponents is the simpler. The following 
sections teach how to use logarithms. 


248. Logarithms of numbers to the base 10 are called common 
logarithms. 

If a number is not an exact power of 10, its logarithm can be 
given only approximately ; a four-place logarithm is one given 
correct to four decimal places. 

Thus the logarithm of 13 is 1.1139; i.e. 13 = 10 11139 , approximately. 












LOGARITHMS 


365 


The integral part of the logarithm is called the characteristic 
and the decimal part, the mantissa. 

The characteristic of log 13 is 1 and the mantissa is .1139. 

Note i. — The plural of mantissa is mantissce. 

Note 2. — A negative number does not have a logarithm. 

249. Finding the characteristic of the logarithm of a number 
greater than 1. 

It is known that 3.53 = 10 5478 , or log 3.53 = .5478. 

Multiplying both members of 3.53 = 10 5478 by 10, 

35.3 = 10- 5478 X 10 1 , or 10 1 - 5478 . log 35.3 = 1.5478. 
Similarly, 353 = 10 1 5478 X 10 1 , or 10 2 - 5478 . /. log 353 = 2.5478. 

The numbers 3.53, 35.3, and 353 have the same significant 
figures; they differ only in the location of the decimal point. 
Their logarithms differ only in their characteristics. These two 
facts indicate a connection between the location of the decimal 
point and the characteristic. 

3.53 has one figure to the left of the decimal point; its logarithm has 
as characteristic 0, which is 1 less than 1. 

35.3 has two figures to the left of the decimal point; its logarithm 
has as characteristic 1, which is 1 less than 2. 

353 has three figures to the left of the decimal point; its logarithm 
has as characteristic 2, which is 1 less than 3. 

Rule. — The characteristic of the common logarithm of a 
number greater than 1 is one less than the number of significant 
figures to the left of the decimal point. 

Thus, the characteristic of log 357.83 is 2; of log 70390.5 is 4. 


EXERCISE 183 

What is the characteristic of the logarithm of: 


1 . 59 

5. 72,860 

9. 5.08 

13. 984.26 

2 . 540 

6 . 11.2 

10 . 30,027.8 

14. 87,600. 

3. 4000 

7. 367.25 

11 . 21.678 

15. 2.1932 

4. 8 

8 . 50900.4 

12 . 100.005 

16. 1,000,000 


366 


ALGEBRA 


Tell the number of significant figures preceding the decimal 
point when the characteristic of the logarithm is: 

17. 5 18. 3 19. 0 20. 1 21. 4 22. 2 

250. Finding the characteristic of the logarithm of a number 
less than 1. Dividing both members of 3.53 = 10- 5478 (§ 249) 
by 10 , 

.353 = 10 5478 -f- 10 1 = 10 5478 ~ 1 . log .353 = .5478 - 1. 

Dividing both members of .353 = 10- 5478-1 , by 10, 

.0353 = 10 5478-1 4- 10 1 = 10- 5478-2 . log .0353 = .5478 - 2. 

Similarly, .00353 = 10- 5478-3 . log .00353 = .5478 - 3. 

Between the decimal point and the first significant figure of: 

.353 there are no zeros ; the characteristic of log .353 is — 1. 

.0353 there is one zero; the characteristic of log .0353 is — 2. 

.00353 there are two zeros; the characteristic of log .00353 is 
- 3. 

Rule. — The characteristic of the common logarithm of a 
(positive) number less than 1 is negative ; numerically it is one 
more than the number of zeros between the decimal point and 
the first significant figure. 

Thus, the characteristic of log .0045 is - 3; of log .00027, is — 4. 


EXERCISE 184 


What is the characteristic of the logarithm of: 


1. .5 

4. .07 

7. .64325 

10. .1007 

2. .004 

5. .01003 

8. .04216 

11. .00008 

3. .00082 

6. .00093 

9. .00205 

12. .24038 


Tell the number of zeros preceding the first significant figure 
when the characteristic of the logarithm is: 

13. - 2 15.-4 17.-1 19. - 5 21.-3 

14. - 8 16. - 10 18. - 6 20.- 7 22.-9 


LOGARITHMS 


367 


251. Another method of writing a negative characteristic. 

In §250 log .353 = .5478 — 1. Actually, therefore, log .353 
is — .4522, a negative number. However, the positive man¬ 
tissa and the negative characteristics are retained. 

.5478 — 1 may be written: 9.5478 — 10. Numerically the 
two expressions have equal value. Note that 9 — 10 = — 1. 

The process in general is to decide upon the characteristic 
by the rule in § 250 ; then, if it is — 1, write it 9 — 10; if — 2, 
write it 8 — 10; etc. 

Thus, log .02 is .3010 - 2, or 8.3010 - 10. 

Note. — The negative characteristic is often written thus: log 02 
= 2.3010; again, log .353 = 1.5478. The minus sign is written over 
the characteristic to indicate that it alone is negative, the mantissa 
being positive. 

EXERCISE 185 

1 -12. Tell how each of the characteristics of the examples 
of Exercise 184 should be written. 

252. Mantissa of a logarithm. From §§ 249 and 250: 

log 3.53 = .5478; log .353 = 9.5478 - 10; 
log 35.3 = 1.5478; log .00353 = 7.5478 - 10. 

The numbers 3.53, 35.3, .353, and .00353 have the same 
significant figures. Their common logarithms have the same 
mantissse. This is an example of the 

Rule. — The common logarithms of all numbers having the 
same significant figures have the same mantissae. 

Thus, the logarithms of 2506, 2.506, 250.6, etc., all have the same 
mantissae. 

253. A table of logarithms consists of the mantissse of the 
logarithms of certain numbers. The characteristics of the 
logarithms are determined by the rules given in §§ 249 and 
250. The table given on pages 370 and 371 gives the man¬ 
tissse of all integers from 100 to 999 inclusive, calculated to 




368 


ALGEBRA 


four decimal places. The decimal point is omitted. Such a 
table is called a four-place table. While a five or six place table 
would be more accurate, this table is sufficiently accurate for 
all ordinary purposes. 

254. Finding the logarithm of a number of three significant 
figures. 

Example 1. —Find the logarithm of 16.8. 

Solution. — 1. In the column headed “No.” find 16. On the hori¬ 
zontal line opposite 16, pass over to the column headed by the figure 8. 
The mantissa 2253 found there, is the required mantissa. 

2. The characteristic is 1, by the rule in § 249. 

3. /. log 16.8 is 1.2253. 

Rule. — To find the logarithm of a number of three figures: 

1. Look in the column headed “ No.” for the first two figures 
of the given number. The mantissa will be found on the hori¬ 
zontal line opposite these two figures and in the column headed 
by the third figure of the given number. 

2. Prefix the characteristic according to §§ 249 and 250. 

Example 2. — Find log .304. 

Solution. — 1. Opposite 30 in the column headed by 4 is the man¬ 
tissa .4829. The characteristic is — 1 or 9 — 10. (§§ 250 and 251.) 

2. /. log .304 = 9.4829 - 10. 

Note. — The logarithm of a number of one or two significant figures 
may be found by using the column headed 0. Thus the mantissa of log 
8.3 is the same as the mantissa of log 8.30; of log 9, the same as of 
log 900. 


EXERCISE 186 


Find the logarithm of: 


1. 365 6. 64 

2. 571 7. 9 

3. 847 8. 5.2 

4. 902 9. 43.6 

5. 200 10. 720 


11. .841 16. .000834 


12 . .0628 17. .07 

13. .00175 18. 3.14 

14. 7680 19. 40.8 

15. 25900 20 . .16 



LOGARITHMS 


369 


255. Finding the logarithm of a number of more than three 
significant figures. 

Example 1. — Find log 327.5. 

Solution. — 1 . From the table • 



2. Since 327.5 is between 327 and 328, its logarithm must be between 
their logarithms. An increase of one unit in the number (from 327 to 
328) produces an increase of .0014 in the mantissa. It is assumed there¬ 
fore that an increase of .5 in the number (from 327 to 327.5) produces 
an increase of .5, of .0014, or of .0007 in the mantissa. 

3. .'. log 327.5 = 2.5145 + .5 X .0014 


= 2.5145 + .0007 = 2.5152. 


This result is obtained in practice as follows. The difference between 
any mantissa and the next higher mantissa as written in the table 
(neglecting the decimal point) is called the tabular difference. The 
tabular difference for this example is 14(5159-5145). .5 of the tabular 

difference is 7. Adding this to 5145 gives 5152, the required mantissa 
of log 327.5. 

Similarly to find log 327.25, the tabular difference is 14. .25 X 14 

= 3.5. Hence the mantissa of log 327.25 is 5145 -f- 3.5 or 5148.5. 
.*. log 327.25 = 2.5149. 

Note 1 . — The process of determining a mantissa which is between 
two mantissae of the table is called interpolation. 

Note 2. — The assumption made in Step 2 is not warranted by the 
facts. Nevertheless, for ordinary purposes, the results obtained in 
this manner are sufficiently correct. 

Note 3. — When interpolating, it is customary to cut down all 
decimals so that the mantissa will again be a four-place decimal. Thus 
3.5 is called 4. 3.4 would be called 3. 

Rule. — To find the logarithm of a number of more than three 
significant figures: 

1. Find the mantissa for the first three figures, and the tabular 
difference for that mantissa. (See Step 3 above.) 

2. Multiply the tabular difference by the remaining figures of 
the given number, preceded by a decimal point. (See Note 3 
above.) 


370 


ALGEBRA 


No. 

0 

1 

2 

3 

4 

5 

0 

7 

8 

9 

IO 

0000 

0043 

0086 

0128 

0170 

0212 

0253 

0294 

0334 

0374 

ii 

0414 

0453 

0492 

0531 

0569 

0607 

0645 

0682 

0719 

0755 

12 

0792 

0828 

0864 

0899 

0934 

0969 

1004 

1038 

1072 

1106 

13 

1139 

“73 

1206 

1239 

1271 

1303 

1335 

1367 

1399 

1430 

14 

1461 

1492 

1523 

1553 

1584 

1614 

1644 

1673 

l 7°3 

1732 

15 

1761 

1790 

1818 

1847 

1875 

1903 

1931 

1959 

1987 

2014 

16 

2041 

2068 

2095 

2122 

2148 

2175 

2201 

2227 

2253 

2279 

17 

2304 

2330 

2355 

2380 

2405 

2430 

2455 

2480 

2504 

2529 

IS 

2553 

2577 

2601 

2625 

2648 

2672 

2695 

2718 

2742 

2765 

19 

2788 

2810 

2833 

2856 

2878 

2900 

2923 

2945 

2967 

2989 

20 

3010 

3032 

3054 

3075 

3096 

3U8 

3i39 

3160 

3181 

3201 

21 

3222 

3243 

3263 

3284 

3304 

3324 

3345 

3365 

3385 

3404 

22 

3424 

3444 

3464 

3483 

3502 

3522 

3541 

3560 

3579 

3598 

23 

3617 

3636 

3655 

3674 

3692 

3711 

3729 

3747 

3766 

3784 

24 

3802 

3820 

3838 

3856 

3874 

3892 

3909 

3927 

3945 

3962 

25 

3979 

3997 

4014 

4031 

4048 

4065 

4082 

4°99 

4116 

4i33 

26 

415° 

4166 

4183 

4200 

4216 

4232 

4249 

4265 

4281 

4298 

27 

4314 

4330 

4346 

4362 

4378 

4393 

4409 

4425 

4440 

4456 

28 

4472 

4487 

4502 

45 1 8 

4533 

4548 

4564 

4579 

4594 

4609 

29 

4624 

4639 

4654 

4669 

4683 

4698 

4713 

4728 

4742 

4757 

30 

4771 

4786 

4800 

4814 

4829 

4843 

4857 

4871 

4886 

4900 

31 

4914 

4928 

4942 

4955 

4969 

4983 

4997 

5011 

5024 

5 0 38 

32 

505! 

5 o6 5 

5°79 

5092 

5105 

5 IJ 9 

5 l 3 2 

5H5 

5*59 

5 I 7 2 

33 

5185 

5^8 

5211 

5224 

5237 

5250 

5263 

5276 

5289 

5302 

34 

5315 

5328 

5340 

5353 

5366 

5378 

5391 

5403 

54i6 

5428 

35 

5441 

5453 

546s 

5478 

5490 

5502 

5514 

5527 

5539 

555i 

36 

5563 

5575 

5587 

5599 

5611 

5 6 23 

5 6 35 

5647 

5658 

5670 

37 

5682 

5694 

5705 

57 1 7 

5729 

5740 

5752 

5763 

5775 

5786 

38 

5798 

5809 

5821 

5832 

5843 

5 8 55 

5866 

5877 

5888 

5899 

39 

59ii 

5922 

5933 

5944 

5955 

5966 

5977 

5988 

5999 

6010 

40 

6021 

6031 

6042 

6053 

6064 

6075 

6085 

6096 

6107 

6117 

4i 

6128 

6138 

6149 

6160 

6170 

6180 

6191 

6201 

6212 

6222 

42 

6232 

6243 

6253 

6263 

6274 

6284 

6294 

6304 

63H 

6325 

43 

6335 

6345 

6355 

6365 

6375 

6385 

6395 

6405 

6415 

6425 

44 

6435 

6444 

6454 

6464 

6474 

6484 

6493 

6503 

6513 

6522 

45 

6532 

6542 

6551 

6561 

6571 

6580 

6590 

6599 

6609 

6618 

46 

6628 

6637 

6646 

6656 

6665 

6675 

6684 

6693 

6702 

6712 

47 

6721 

6730 

6739 

6749 

6758 

6767 

6776 

6785 

6794 

6803 

48 

6812 

6821 

6830 

6839 

6848 

6857 

6866 

6875 

6884 

6893 

49 

6902 

6911 

6920 

6928 

6937 

6946 

6 955 

6964 

6972 

6981 

50 

6990 

6998 

7007 

7016 

7° 2 4 

7033 

7°4 2 

7050 

7059 

7067 

5i 

7076 

7084 

7093 

7101 

7110 

7118 

7126 

7135 

7H3 

7152 

52 

7160 

7168 

7177 

7185 

7*93 

7202 

7210 

7218 

7226 

7235 

53 

7243 

7251 

7259 

7267 

7275 

7284 

7292 

7300 

7308 

73i6 

54 

7324 

7332 

7340 

7348 

735 6 

7364 

7372 

738o 

7388 

7396 

No. 

0 

1 

2 

3 

4 

5 

6 

7 

8 

9 



















LOGARITHMS 


371 


No. 

0 

1 

2 

3 

4 

5 

6 

7 

8 

9 

55 

7404 

7412 

74 i 9 

7427 

7435 

7443 

7451 

7459 

7466 

7474 

56 

7482 

7490 

7497 

75°5 

75 1 3 

7520 

7528 

7536 

7543 

755 1 

57 

7559 

7566 

7574 

7582 

7589 

7597 

7 6 °4 

7612 

7619 

7627 

58 

7634 

7642 

7649 

7657 

7664 

7672 

7679 

7686 

7694 

7701 

59 

7709 

rji6 

7723 

773 i 

7738 

7745 

775 2 

7760 

7767 

7774 

60 

7782 

7789 

7796 

7803 

7810 

7818 

7825 

7832 

7839 

7846 

61 

7853 

7860 

7868 

7875 

7882 

7889 

7896 

7903 

7910 

7917 

62 

7924 

7931 

7938 

7945 

7952 

7959 

7966 

7973 

7980 

7987 

63 

7993 

8000 

8007 

8014 

8021 

8028 

8035 

8041 

8048 

8055 

64 

8062 

8069 

8075 

8082 

8089 

8096 

8102 

8109 

8116 

8122 

65 

8129 

8136 

8142 

8149 

8156 

8162 

8169 

8176 

8182 

8189 

66 

8195 

8202 

8209 

8215 

8222 

8228 

‘8235 

8241 

8248 

8254 

67 

8261 

8267 

8274 

8280 

8287 

8293 

8299 

8306 

8312 

8319 

68 

8325 

8331 

8338 

8344 

8351 

8357 

8363 

837 ° 

8376 

8382 

69 

8388 

839s 

8401 

8407 

8414 

8420 

8426 

8432 

8439 

8445 

70 

845 1 

8457 

8463 

8470 

8476 

8482 

8488 

8494 

8500 

8506 

71 

8513 

8519 

8525 

8531 

8537 

8543 

8549 

8555 

8561 

8567 

72 

8573 

8579 

8585 

8591 

8597 

8603 

8609 

8615 

8621 

8627 

73 

8633 

8639 

8645 

8651 

8657 

8663 

8669 

8675 

8681 

8686 

74 

8692 

8698 

8704 

8710 

8716 

8722 

8727 

8733 

8739 

8745 

75 

8751 

8756 

8762 

8768 

8774 

8779 

8785 

8791 

8797 

8802 

76 

8808 

8814 

8820 

8825 

8831 

8837 

8842 

8848 

8854 

8859 

77 

8865 

8871 

8876 

8882 

8887 

8893 

8899 

8904 

8910 

8915 

78 

8921 

8927 

8932 

8938 

8943 

8949 

8954 

8960 

8965 

8971 

79 

8976 

8982 

8987 

8993 

8998 

9004 

9009 

9015 

9020 

9025 

80 

9031 

9036 

9042 

9047 

9053 

9058 

9063 

9069 

9074 

9079 

81 

9085 

9090 

9096 

9101 

9106 

9112 

9117 

9122 

9128 

9133 

82 

9138 

9143 

9149 

9154 

9159 

9165 

9170 

9175 

9180 

9186 

83 

9191 

9196 

9201 

9206 

9212 

9217 

9222 

9227 

9232 

9238 

84 

9243 

9248 

9253 

9258 

9263 

9269 

9274 

9279 

9284 

9289 

85 

9294 

9299 

9304 

9309 

9315 

9320 

9325 

9330 

9335 

9340 

86 

9345 

935 ° 

9355 

9360 

9365 

937 ° 

9375 

9380 

9385 

9390 

87 

9395 

9400 

9405 

9410 

94 i 5 

9420 

9425 

9430 

9435 

9440 

88 

9445 

945 ° 

9455 

9460 

9465 

9469 

9474 

9479 

9484 

9489 

89 

9494 

9499 

95°4 

95°9 

95*3 

95 l8 

9523 

9528 

9533 

9538 

90 

9542 

9547 

9552 

9557 

9562 

9566 

9571 

9576 

958 i 

9586 

9i 

9590 

9595 

9600 

9605 

9609 

9614 

9619 

9624 

9628 

9633 

92 

9638 

9643 

9647 

9652 

9657 

9661 

9666 

9671 

9675 

9680 

93 

9685 

9689 

9694 

9699 

97°3 

97°8 

9713 

9717 

9722 

9727 

94 

9731 

9736 

9741 

9745 

975 ° 

9754 

9759 

9763 

9768 

9773 

95 

9777 

9782 

9786 

9791 

9795 

9800 

9805 

9809 

9814 

.9818 

96 

9823 

9827 

9832 

9836 

9841 

9845 

9850 

9854 

9859 

9863 

97 

9868 

9872 

9877 

9881 

9886 

9890 

9894 

9899 

9903 

9908 

98 

9912 

9917 

9921 

9926 

9930 

9934 

9939 

9943 

9948 

9952 

99 

9956 

9961 

9965 

9969 

9974 

9978 

9983 

9987 

9991 

9996 

No. 

0 

1 

2 

3 

4 

5 

0 

7 

8 

9 




















372 


ALGEBRA 


3. Add the result of Step 2 to the mantissa obtained in Step 1. 

4 . Prefix the proper characteristic, determined by §§ 249 and 
• 250. 

Example 2. — Find log 34.652. 

Solution. — 1. Mantissa of log 346 = 5391 
Mantissa of log 347 = 5403 

2. Tabular difference = 12. .52 X 12 = 6.24, or 6 

3. .*. Mantissa for log 34652 = 5391 + 6, or 5397 

4. Y. log 34.652 = 1.5397. 

Example 3. —Find log .021508. 

Solution. — 1. Mantissa of log 215 = 3324 
Mantissa of log 216 = 3345 

2. Tabular difference = 21. .08 X 21 = 1.68, or 2 

3. mantissa of log 21508 = 3324 + 2, or 3326 

4. .*. log .021508 = .3326 - 2, or 8.3326 - 10. 

EXERCISE 187 

What is the tabular difference when the mantissa is: 

1. 2380 3. 5752 5. 8949 7. 9595 9. 5237 

2 . 3243 4. 7427 6 . 4771 8 . 1461 10 . 7774 


Find the logarithm of: 


11. 

342.5 

16. 501.6 

21. .04873 

26. 

1.0673 

12. 

252.1 

17. 28.25 

22. 328.15 

27. 

.052425 

13. 

865.2 

18. 1.158 

23. 453.24 

28. 

3.1416 

14. 

764.4 

19. 7.631 

24. 86.431 

29. 

6324.5 

15. 

438.3 

20 . .5842 

25. 3.7275 

30. 

.008004 


256. Finding the number corresponding to a given logarithm. 

Example 1. — Find the number whose logarithm is 1.6571. 

Solution. — 1. Find the mantissa 6571 in the table. 

2. In the column headed “No.” on the line with 6571 is 45. These 
are the first two figures of the number. At the head of the column con¬ 
taining 6571 is 4, the third figure of the number. Hence the number 
sought has the figures 454. 



LOGARITHMS 


373 


3. The characteristic being 1, the number must have two figures to 
the left of the decimal point. (§ 249) 

4. the number is 45.4. 


Rule. — To find the number corresponding to a given loga¬ 
rithm when the mantissa appears in the table : 

1. Find the three figures corresponding to this mantissa, as 
in Example 1. 

2 . Place the decimal point according to the rules in §§ 249 
and 250. 

EXERCISE 188 


Find the number whose logarithm is : 


1. 

3.4786 

6. 3.8615 

9. 

9.2405 - 

10 

13. 

2 . 

1.7300 

6 . 0.6284 

10 . 

8.7966 - 

10 

14. 

3. 

.9325 

7. 1.3962 

11 . 

7.9868 - 

10 

15. 

4. 

2.2095 

8 . 2.5105 

12 . 

9.4728 - 

10 

16. 


Example 2. — Find the number whose logarithm is 


5.9149 

3.6776 

4.8899 

1.3010 

1.3934. 


Solution. — 1. The mantissa 3934 does not appear in the table. 

The next less mantissa is 3927, and the next greater is 3945. 

The corresponding numbers are 247 and 248. That is: 
mantissa of log 247 = 3927 1 Diff. ) Tabular 
mantissa of log x = 3934 / = 7. } difference 
mantissa of log 248 = 3945 j = 18. 

2. Since an increase of 18 in the mantissa produces an increase of 1 
in the number, it is assumed that an increase of 7 in the mantissa must 
produce an increase of ^ or .38 in the number. Hence the number has 
the figures 247.38. 

3. Since the characteristic is 1, the number must be 24.738. 


Rule. — To find the number corresponding to a given log¬ 
arithm when the mantissa does not appear in the table : 

1. Find in the table the next less mantissa. Find the corre¬ 
sponding number of three figures, and the tabular difference. 

2. Subtract the next less mantissa from the given mantissa, 
and divide the remainder by the tabular difference to two deci¬ 
mal places. 


374 


ALGEBRA 


3. Annex the quotient to the number of three figures obtained 
in Step 1. 

4. Place the decimal point according to the rules in §§ 249 
and 250. 

EXERCISE 189 


Find the number whose logarithm is: 


1 . 

2.7408 

6. 

9.5969-10 

11. 2.2930 

16. 

8.9194-10 

2. 

1.6678 

7. 

8.3429-10 

12. 1.9665 

17. 

7.7978-10 

3. 

.4188 

8. 

3.8497 

13. 3.8598 

18. 

3.2306 

4. 

3.8983 

9. 

7.3288-l0 

14. 9.7606-10 

19. 

1.8817 

5. 

2.9417 

10. 

.1195 

16. 4.3346 

20. 

2.9986 


PROPERTIES OF LOGARITHMS 

257. The preceding discussion relates entirely to the com¬ 
mon system of logarithms (§ 248). Certain properties of log¬ 
arithms to any base will be considered now. 

Note. — The base may be any positive number different from 1. 

258. Just as logi 0 3.053 = .4847 means that 10 4847 = 3.053, 
so loga N = x means that N = a x . 

Loga N is read “ the logarithm of N to the base a.” 

259. Logarithm of a product. 

Assume that a x = M 1 \x = log a M, 

and a y = N ] n \y = \og a N. 

Also a x ' a v = MN, or a x+ * = MN. log 0 MN = x + y. 

(§ 258) 

Therefore log a MN = log a M + log a N. 

Rule. — In any system, the logarithm of a product is equal 
to the sum of the logarithms of its factors. 

Example. — Given log 2 = .3010, and log 3 = .4771, find 
log 72. 


<© S’ 


LOGARITHMS 


375 


Solution. —1 . log 72 = log 2 • 2 • 2 • 3 • 3 

2. = log 2 + log 2 + log 2 + log 3 + log 3 

3. log 72 = 3 log 2 -f 2 log 3, or 3(.3010) + 2(.4771) 

= .9030 + -9542, or 1.8572. 

EXERCISE 190 

Given log 2 = .3010, log 3 = .4771, and log 5 = .6990. 
Find the following logarithms as above; check the solutions 
by finding the same logarithms in the table: 

1. log 15 4. log 45 7. log 144 10. log 240 

2. log 30 6. log 36 8. log 256 11. log 432 

3. log 60 6. log 150 9. log 360 12. log 375 


13. Find by logarithms the value of 35.2 X 2.35 X 6.43. 

Solution. — 1. Let v = 35.2 X 2.35 X 6.43 log 35.2 = 1.5465 

2. log t; = log 35.2 + log 2.35 + log 6.43 log 2.35 = 0.3711 


3. .’. log v = 2.7258 

4. .*. v = 531.87 (§ 256). 

Find by logarithms the value 

14. 42.5 X 26.8 17. 239.5 

15. 3.89 X 72.6 18. 871.2 

16. 535 X .621 19. 1.414 


log 6.43 = 0.8082 
2.7258 


of: 


X 38.4 

20. 7.026 X 8059 

X 45 

21. 432.4 X 1.658 

X 360 

22. 93.62 X 768.75 


23. Find by logarithms the value of .0631 X 7.208 X .51272. 
Solution. — 1. log v = log .0631 + log 7.208 + log .51272 


2 . 


3. 


log .0631 = 8.8000 - 10 

log 7.208 = 0.8578 

log .51272 = 9.7099 - 10 


.-. log v = 9.3677 - 10 


19.3677 - 20 = 9.3677 - 10 
/. v = .2332. 


Note. — If the sum of the logarithms is a negative number, the 
result should be written so that the negative part of the characteristic 
— 10. In this example, the characteristic 19 — 20 is changed to 
- 10 . 






376 


ALGEBRA 


Find by logarithms the value of: 

24. .342 X 2.15 27. 76.85 X .00435 

25. .6546 X .0368 28. .0169 X .347 X .846 

26. .7932 X 8.421 29. .081 X .3806 X 563 

260. Logarithm of a quotient. 

Assume that a x = M 1 , f x = log a M, 

and a v = N \ * \y = log a N. 

Also, a x + a v = M -i- N, or a x ~ v = M -r- N. 

loga (M -s- N) = x — y. 

Therefore, log a (M + N) = logo M — log a N. 

Rule. — In any system, the logarithm of the quotient of two 
numbers is equal to the logarithm of the dividend minus the 
logarithm of the divisor. 

Example 1. — Given log 2 = .3010 and log 3 = .4771, find 
log f. 

Solution. — 1! log f = log 3 — log 2 = .4771 — .3010 = .1761. 

Example 2. — Find log f. 

2-2-2 

Solution. — 1. log f = log —-—— 

O * O 

2. = (log 2 + log 2 + log 2) - (log 3 + log 3) 

3. = 3(.3010) - 2(.4771) = .9030 - .9542 

4. .-. log | = 9.9488 - 10. 

Note 1. — To subtract a logarithm from a smaller logarithm, or to 
subtract a negative logarithm from a positive one, increase the char¬ 
acteristic of the minuend by 10, writing — 10 after the mantissa to 
compensate. Thus, in this example, .9542 is greater than .9030; 
therefore, .9030 is written 10.9030 — 10, after which the subtraction 
is performed. 

Note 2. — To find the logarithm of a fraction, add the logarithms 
of the factors ’of the numerator, and from the result subtract the sum 
of the logarithms of the factors of the denominator. 


.9030 = 10.9030 - 10 


.9542 = 


.9542 


9.9488 - 10 





LOGARITHMS 


377 


EXERCISE 191 

Given log 2 = .3010, log 3 = .4771, and log 5 = .6990. 

Find: 


1. logf 3. log ^ 5. log ff 7. logf 

2. log^- 4. logJ;2 6. log^ 8. log 

Find by logarithms the value of: 


9. 

335 - 5 - 56 

12. 

230.4 4 - 

125 

15. 3305 

1.414 

10. 

483 -h 71 

13. 

739.75 - 

4 1.73 16. 8.964 -h 

45.25 

11. 

688 49 

14. 

400.36 ■ 

4 2.53 1.7. 6.429 4 - 

56.84 

18. 

4.16 X 32 



22. 

3.25 X .0063 



485 




.007 


19. 

35.2 X 1.52 



23. 

527.8 X .069 


53.87 




2.449 


20. 

43.57 X .069 



24. 

4.501 X 837 X .0239 

3.14 




256 X 324 


21. 

14.07 X 347 



25. 

302 X 1.414 X .0293 

18 




450 X .3645 



261. The logarithm of a power of a number. 

Assume that a x = M ; or x = log a M. 

Then, ( a x ) p = M p , or a px = M p . log a M p = px 
Therefore, logo M p = p logo M. 

Rule. — In any system, the logarithm of any power of a 
number is equal to the logarithm of the number multiplied by 
the exponent indicating the power. 

Example 1. — Given log 7 = .8451, find log 7 5 . 

Solution . — log 7 5 = 5 log 7 = 5 X .8451 = 4.2255 
Example 2. — Find by logarithms 1.04 10 . 

Solution . — 1. log 1.04 10 = 10 log 1.04 = 10 X .0170 = .1700 

2. The number whose logarithm is .1700 is 1.479. (§ 256) 

3. /. 1.04 10 = 1.479. 










378 


ALGEBRA 


Example 3. — Find by logarithms >^365. 

Solution. — 1. log >^365 = log 365^ = £ log 365 

2. log v/365 = £ X 2.5623 = 0.8541 

3. The number whose logarithm is 0.8541 is 7.146. (§256) 

4. <^365 = 7.146. 

When finding a cube root, the logarithm of the radicand is 
divided by 3; when finding a square root, the logarithm of the 
radicand is divided by 2. This suggests the 

Rule. — In any system, the logarithm of a root of a number 
is the logarithm of the radicand divided by the index of the root. 

Example 4. — Find by logarithms ^.0359. 

Solution. — 1. log ^.0359 = £ log .0359 = £(8.5551 — 10) 

2 . log v^.0359 — £(38.5551 — 40) (See note below.) 

3. .*. log 0359 = 9.6387 - 10 

4. The number whose logarithm is 9.6387 — 10 is .4352. (§ 256) 

5. v^.0359 = .4352. 

Note. — To divide a negative logarithm, write it in such form that 
the negative part of the characteristic may be divided exactly by the 
divisor, and give —10 as quotient. 

Thus, 8.5551 — 10 is changed to 38.5551 — 40 since the divisor is 4. 
If the divisor were 3, it would be changed to 28.5551 — 30. 


EXERCISE 192 


Given log 2 
Find: 


.3010, log 3 = .4771, and log 5 


.6990. 


1. log 3 7 3. log 5 4 

2. log 3 5 4. log 15 s 


6. log (30)* 7. log V / 10 

6. log V5 8. log 


Find by logarithms the value of: 


9. 323 2 

13. 

10. 4.025 2 

14. 

11. V418.5 

15. 

12. V784 

16. 


V92.04 

8.975 2 

i X 3.1416 X 6.5 3 
3.1416 X 19 2 








LOGARITHMS 


379 


17. JM 

^ 894.5 

18. V32.2 X 20.5 X 12.8 


19. X 



21. The volume of a right circular cylinder is given by the 
formula V = ttR 2 H. Find V by logarithms: 

a. if R = 12.4 and H = 30.3. b. if R = 7.6 and H = 28.5. 

22. The distance covered by a freely falling body is given by 
the formula s = %gt 2 . Find s if g = 32.16 feet and t = 4.5 
seconds. 

23. The volume of a cone is given by the formula V = 7 rR 2 H. 

Find V if R = 9.5 and H = 22. 

24. The volume of a sphere is given by the formula V = 

i ttR\ Find V : a. if R = 5.4. 6. if R = 10.5. 

25. The simple interest formula is I = PRT. Find I: 

a. if P = $845-; R = 4%; and T = 5.5 yr. 

b. if P = $1850; R = 3.5%; and T = 7 yr. 9 mo. 

26. The amount to which P dollars will accumulate at R % 
compounded annually for n years is given by the formula 

- f (> + !§)"• 

а. if P = $450; R = 4 %; and n = 8 yr. 

б. if P = $125; R = 3.5% ; and w = 20 yr. 

27. The area of an equilateral triangle whose side is s inches 

long is given by the formula A = -^a/ 3. Find A when s = 12. 

28. Using the formula F = —, find F when m = 150, g = 32, 

gr 

v = 25, and r = 5. 

29. Using the formula Z = 2 7 rrA, find Z when tt = - 2 /, r = 
11.2, and h = 3. 

30. Using the formula S = ^ l • 7r(r + find S when l = 9.7, 
r — 3.6, and 5 = 5.7. 





XXII. PROGRESSIONS 


ARITHMETIC PROGRESSION 


262. An arithmetic progression (A. P.) is a sequence of 
numbers, called terms, each of which after the first is derived 
from the preceding by adding to it a fixed number, called the 
common difference. 

Thus, 1, 3, 5, 7, — is an A. P. Each term after the first is derived 
from the preceding by adding 2. The next two terms are 9 and 11. 
2 is the common difference. 

Again, 9, 6, 3, •••• is an A. P. The common difference is — 3. The 
next two terms are 0 and — 3. 

Note. — The common difference is found by subtracting any term 
from the one following it. 


EXERCISE 193 


Determine which of the following are arithmetic progressions ; 
determine the common difference and the next two terms of 
the arithmetic progressions: 


1. 5, 8, 11, 14, •••• 

2. 3, 8, 13, 18, •••• 

3. 1, 5, 8, 13, •••• 

4. 32, 26, 20, 14, • 


5. If, 2±, 3, 3f, 

6. 6 r, 8.5 r, 11 r, .... 

7. 3 a, .5 a, — 2 a, •••• 

8. 1.05, 1.10, 1.15 


9. c + d, 2 c + d, 3 c + d 

10 . 6 m + 8 n, 4 m -f 9 n, 2 m + 10 n, •••• 


Write the first five terms of the A. P. 

11 

12 

13 

14 

16 

in which the first term is ... 

13 

22 

3.5 

y 

a 

and the common difference is 

6 

- 7 

4.5 

- 5 

d 


16. Find the 15th term of the A. P. whose first term is 5, 
and common difference 7. 

380 













ARITHMETIC PROGRESSION 


381 


263. The nth term of an arithmetic progression. It is 

possible to determine a particular term of an arithmetic pro¬ 
gression without finding all of the preceding terms. 

Given — the first term a , the common difference d, and the 
number of the term, n, of an arithmetic progression. 


Find — the nth term, l. 

Solution. — 1. The progression is 


Term 1 

Term 2 

Term 3 

Term 4 

Term 10 

Term n 

a 

a -f d 

a -j- 2 d 

a -f- 3 d 

....? 

.... ? 


2. The coefficient of d in each term is 1 less than the number 
of the term. 

3. What is the 10th term? the 18th? the 35th? 

4. Similarly, the coefficient of d in the nth term is (n — 1). 

• 


5. 


I = a + (n - l)d 


Example. —Find the 10th term of 8, 5, 2, •••*. 

Solution. — 1. This is an A. P. in which a = 8, d = — 3, n = 10, and 
l = ? 

2. The formula is l = a + (n — 1 )d. 

3. /. I = 8 + 9(— 3) = 8 - 27, or - 19. 


EXERCISE 194 

Find: 

1. The 10th term of 4, 10, 16, —; also the 18th. 

2. The 12th term of 17, 14, 11, ••••; also the 26th. 

3. The 16th term of — 3, — 8, — 13, —; also the 36th. 

4. The 21st term of 2, 2.05, 2.10, ••••; also the 46th. 

5. The 11th term of 1, l-J, 2, ••••; also the 52d. 

6. What term of the progression 1.05, 1.10, 1.15, •••• is 2.25? 


Solution. — 1. This is an A. P. in which a = 1.05, 
and l = 2.25. The formula is l = a + (n — 1 )d. 

2. .*. 2.25 - 1.05 + (n - 1) • .05. 

(Complete the solution.) 



? 

















382 


ALGEBRA 


7. What term of the progression 7, 4, 1, •••• is — 80? 

8. What term of the progression — 51, — 43, — 35, ••••is 
205? 

9. What term of the progression -J, f, ••••is 23£? 

10. What term of the progression 1.07, 1.14,1 .21, •••• is 3.17? 

Find the common difference : 

11. If the first term of an A. P. is 6 and the 27th term is 188. 

12. If the first term is 8 and the 21st term is 108. 

13. If the first term is — 25 and the 23d term is — 14. 

14. If the first term is 430 and the 41st term is 110. 

15. Find the 12th term of the A. P. whose first term is 10 and 
whose 19th term is 100. 

16. A man is paying for a house on the installment plan. His 
payments during the first three months are $20.00, $20.10, and 
$20.20. What will his 20th and 30th payments be ? 

17. The first three payments on a washing machine are $8.00, 
$8.04, and $8.08. What will the 12th payment be ? 

18. In a Christmas savings fund, the payments to be made 
for 50 weeks are 5|£, 10^, 15^, etc. What will the 40th payment 
be ? the 50th ? 

19. Another -savings fund plan calls for payments for 50 
weeks as follows : $5.00, $4.90, $4.80, etc. What will the 25th 
payment be ? the 50th ? 

264. The terms of an arithmetic progression between any 
two other terms are called the arithmetic means of these two 
terms. 

Thus, the three arithmetic means of 2 and 14 are 5, 8, 11, since 2, 5, 
8, 11, 14 form an arithmetic progression. 

A single arithmetic mean of two numbers is particularly 
important. It is called the arithmetic mean of the numbers. 


ARITHMETIC PROGRESSION 


383 


When two numbers are given, any specified number of arith¬ 
metic means can be inserted between them. 

Example. — Insert five arithmetic means between 13 and 

- 11 . 

Solution. — 1. There will be an arithmetic progression of 7 terms, 
in which a = 13, l = — 11, and n = 7. Find d. 

2. 1 = a + (n - 1 )d. - 11 = 13 + 6 d, or d = - 4. 

3. The progression is: 13, 9, 5, 1, — 3, — 7, — 11. 

Check. — This is an A. P. with five terms between 13 and — 11. 

EXERCISE 195 

1. Insert three arithmetic means between 1 and 17. 

2. Insert four arithmetic means between — 14 and 16. 

3. Insert seven arithmetic means between 8 and 20. 

4. Insert five arithmetic means between 1^ and 6. 

5. Insert four arithmetic means between — f and — 5. 

6. Find the arithmetic mean of 5 and 13. 

7. Find the arithmetic mean of ( y — 4) and ( y +4). 

8. Find the arithmetic mean of V3 and V27. 

9. Find the arithmetic mean of a and h. From the result, 
make a rule for finding the arithmetic mean of any two numbers. 

Note. — The arithmetic mean of two numbers is also called their 
average. 

10. a. Find the arithmetic mean of 11 and 19. 

h. Check your results by finding this arithmetic mean by 
use of the formula derived in Example 9. 

11. Find the common difference if 3 arithmetic means are in¬ 
serted between g and h. 

12. Find the common difference if r arithmetic means are 
inserted between 5 and 45. 

13. Find the common difference if r arithmetic means are in¬ 
serted between s and t. 


384 


ALGEBRA 


265. The sum of the first n terms of an arithmetic progression 

can be found without writing down all the terms. 

Given — the first term, a, the number of terms, n, and the 
nth term, l. 


Find — the sum of these n terms, S. 

Solution. — 1 . S = a + (a + d) + (a + 2 d) • • • -f- {l — 2 d) -f- (l — d)-\- l. 

2. Writing the terms in reverse order, 

S = l-\- (l — d) + (Z — 2 d) •••-!- (a -1- 2 d) + (<z-|-d) -f- n. 

3. Adding the equations in Steps 1 and 2, 

2 S = (a + l) + (a + l) + (a + l) • • • + (a ,+ l) + (a + l) + (a + 1). 

4. Since there were n terms in equations 1 ]and 2, and since a sum 
(a + l) results in Step 3 from each term of equation 1, 

2S = n(a + l). 


5. 


s = | (« + 0 


6. By § 263, l = a + (n — l)d. Substituting this value of l in Step 5, 


7. 


S = 5lo + {o + (»-l)da. 

A 

\ S = - J2 a + ( n- l)d| 
2 


Note. — The formula} in Steps 5 and 7 should be memorized. 


Example. — Find the sum of the first 12 terms of the pro¬ 
gression 8, 5, 2, •••• 

Solution. — 1. a = 8; d = — 3; n = 12; S = ? 

2. S = |{2 a + (n - 1 )d}. S = 6{16 + 11 • (- 3)}. 

3. S = 6(— 17) = - 102. 


EXERCISE 196 

Find the sum of: 

1. 14 terms of 4, 9, 14. 

2. 18 terms of 6, 1, — 4. 

3. 15 terms of — 80, — 71, — 62. 

4. 12 terms of $1.08, $1.16, $1.24. 




ARITHMETIC PROGRESSION 


385 


Find the sum of the terms of an arithmetic progression if: 

5. The number is 16, the first is 3, and the last is 48. 

6. The number is 39, the first is 0, and the last is 50. 

7. The number is 19, the first is 24, and the last is - 48. 

8. The number is 6, the first is — f, and the last is 

9. Find the sum of the even numbers from 2 to 50. 

10. Find the sum of the numbers 3, 6, 9, •••• 99. 

11. Find the sum of the odd numbers from 1 to 199. 

12. Find the sum of all the even integers beginning with 2 
and ending with 400. 

13. A pile of fence posts has 40 in the first layer, 39 in the 
second, 38 in the third. There are 20 layers. How many 
posts are there in the pile? 

14. In a certain school system, a teacher is paid $950 for her 
first year’s work, and is given an increase of $50 per year each 
year thereafter. What will be the teacher’s total income during 
ten years of service ? 

15. On a debt of $4000, a man was paying 7% interest. At 
the end of each year, he plans to pay $200 on the principal, 
and also the interest on the balance of the debt during that year. 

a. Write down his payments to be made at the end of years 
one, two, and three. 

b. How many years will it take him to pay off the debt? 

c. How much will he have paid by the time he has freed him¬ 
self of the debt? 

16. On a large apartment building there is a debt of $75,000, 
with interest at 6%, payable at the end of each year. Each 
year, $5000 of the debt is paid off, thus decreasing both the 
debt and the interest to be paid each year. 

a. What is the payment due at the end of years one, two, 
and three ? 

b. How much will have been paid for interest by the time the 
debt is paid off? 


386 


ALGEBRA 


17. It has been learned that if a marble, placed in a groove 
on an inclined plane , passes over a distance D in one second, 
then in the second second it will pass over the distance 3 D, 
in the third, over the distance 5 D, etc. Over what distance 
will it pass in the 10th second ? in the tt h second ? 

18. Through what total distance does the marble in Example 
17 pass in 5 seconds? in 10 seconds? in t seconds? 

19. Experiment has shown that an object will fall during 
successive seconds the following distances : 

1st second, 16.08 ft.; 3d second, 80.40 ft.; 

2d second, 48.24 ft.; 4th second, 112.56 ft. 

Find the distance through which the object will fall during 
the 7th second; the fth second. 

20. Find the total distance through which the object in 
Example 19 falls in 5 seconds; in t seconds. 

21. Substitute g for 32.16 in the final result of Example 20 
and simplify the result. 

SUPPLEMENTARY TOPIC 

266. In an arithmetic progression, there are five elements, 
a, d, l, n, S. Two independent formulse connect these elements, 
the formula for the sum and the formula for the term 1. Hence, 
if any three of the elements are known, the other two may be 
found. 

Note. — Remember that the formula for the sum is given in two 
ways. 

Example 1. — Given a = — -J, n = 20, S = — ■find d 
and l. 

Solution. — 

1. S = | (a + l ). .*. - | = 10 ( — | + i'j ; whence l = |- 

2. I = a + (n — 1 )d. .*. ^ = — § + (19) • d ; whence d = -• 

Z 6 6 


ARITHMETIC PROGRESSION 


387 


Example 2. — Given a = 7, d = 4, S = 403; find n and l. 
Solution. — 

1. S = |{2o + (» - l)d}. 403 = 2(14 + (n - 1) • 4!. 

2. /. 806 = n{4 n + 10}; 4 n 2 + 10 n - 806 = 0; 

2 rc 2 + 5 n - 403 = 0. 

3 . n _ ~ 5 =*= V25 + 3224 _ - 5 =±= V3249 

4 4 

= ~ 5 =*= 57 

:.n=- or + 13. 

4 

Since rc, the number of terms, must be an integer, n must be 13, and 
not — 

4. l = 7 + 12 • 4 = 55. 

Note. — A negative or a fractional value on n must be rejected, 
together with all other values depending upon it. 

Example 3. — The sixth term of an arithmetic progression is 
10 and the 16th term is 40. Find the 10th term. 

Solution. — 1. By the formula l = a -f (n — 1 )d : 

o + 5d =10. 
a + 15 d = 40. 

2. Solving the system of equations in Step 1, d = 3 and a = —5. 

3. The 10th term: l = - 5 + 9*3 = - 5+27 = 22. 

EXERCISE 197 

1. Given l = 57, d = 4, n = 17; find a and S. 

2. Given a = —7,1 = 123, n = 27; find d and S. 

3. Given a = f, l = S = ; find n and d. 

4. Given a = f, / = — -§, d = — i; find n and S. 

6. Given a = -f, n = 16, S = find Z and d. 

6. Given n = 19, S = — cZ = ; find a and Z. 

7. Given a = — -J, Z = — S = — 143; find n and cZ. 

8. Given a = d = — f, S = — 2; find w and Z. 

9. Given a, Z, and cZ; derive a formula for n. 

10. Given a, , and n ; derive a formula for <Z. 






388 


ALGEBRA 


11. Given a, n, and S ; derive a formula for l. 

12. Given d, n, and S ; derive a formula for a. 

13. Given d, l, and n ; derive a formula for a and for S. 

14. The 6th term of an arithmetic progression is 25 and the 
11th term is 5. Find the 30th term. 

15. The 5th term of an arithmetic progression is , the 21st 
term is and the last term is . Find the number of the 
terms. 

16. The sum of the 2d and the 9th terms of an arithmetic 
progression is — 8; and the sum of the 5th and the 10th terms 
is — f. Find the first term. 

17. Find four numbers in arithmetic progression such that 
the sum of the first two shall be — 51, and the sum of the last 
two shall be 9. 

18. Find five numbers in arithmetic progression such that 
the sum of the second, third, and fifth shall be 7; and such 
that the product of the first and fourth shall be — 14. 

19. Find three numbers in arithmetic progression such that 
the sum of the squares of the first two shall be 193; and such 
that the sum of the first and third shall be 7 less than 3 times 
the second. 

20. Find three integers in arithmetic progression such that 
their sum shall be 6, and their product — 192. 

GEOMETRIC PROGRESSION 

267. A geometric progression (G. P.) is a sequence of numbers 
called terms, each of which, after the first, is derived by multi¬ 
plying the preceding term by a fixed number called the ratio. 

Thus, 2, 6, 18, 54, • • • • is a geometric progression. Each term, after the 
first, is obtained by multiplying the preceding term by 3. The ratio is 3. 

Again, 15, — 5, + f, — f, •••• is a G. P. The ratio is — The 
next two terms are + ^ and — 

Note. — The ratio may be found by dividing any term by the one 
preceding it. 


GEOMETRIC PROGRESSION 


389 


EXERCISE 198 


Determine which of the following are geometric progressions; 
determine the ratio and also the next two terms of the geometric 
progressions: 


1. 2, 4, 8, 16, •••• 

2. 64, 32, 16, — • 

3. 81, 30, 21, •••• 


4. 3, - 6, 12, - 24, 
7 n In 


6. 7 n, 


9 ’ 


6. 2y, 6 y\ 18 y\ •••• 

7. (m + n), (m + n) 2 , (m + n) 3 , •••• 


8. 3, - 9, - 27, 81 


9. 


1 

—) 

a 


a 3 ’ 


1 



5 

r 2 ’ 


_ 5 _ 

3 r 2> 


Write the first five terms of the G. P. in which 



11 

12 

13 

14 

15 

The first term is 

- 3 

72 

i 

4 

iv 

a 

The ratio is 

- 2 

i 

4 

3 


r 


268. The nth term of a geometric progression can be found 
without determining all the preceding terms. 

Given — the first term, a ; the ratio, r ; and the number of 
terms, n, of a geometric progression. 

Find — the nth term, l. 

Solution. — 1. The progression is a, ar, ar 2 , •••• 

2. The exponent of r in each term is 1 less than the number 
of the term. Hence the 8th term would be ar 7 and the 20th 
term, ar 19 . 

3. Similarly, the nth term must be ar n ~ l . 

I = ar "- 1 











390 


ALGEBRA 


EXERCISE 199 

1. Find the 8th term of 1, 2, 4, •••• 

2. Find the 6th term of 5, 3, §, •••• 

3. Find the 11th term of 4, — 12, 36, •••• 

4. Find the 7th term of — f-, 3, — 6, •••• 

6. Find the 9th term of 5, f ) f, •••• 

6. Find the 10th term of , —, •••• 

128 64’ 32 

7. Indicate the 12th term of 1, (a + b), (a + 6) 2 , •••• 

8. Indicate the 16th term of 1, ••••; also the tth term. 

9. Indicate the 12th term of x, —, —, —, ••••; also the 

4 16 64 

(n + l)th term. 

10. What term of the progression 5, 10, 20, 40, ••••is 640? 

11. What term of the progression 7, 14, 28, •••• is 448? 

12. What term of the progression 12, 6, 3, •••• is ^? 

13. If the first term of a geometric progression is 4 and the 
5th term is fa, what is the ratio ? 

Find the ratio of the geometric progression if: 

14. The first term is 4 and the 5th term is fa. 

16. The first term is f- and the 6th term is 54. 

Find by logarithms the value of: 

16. The 15th term of the G. P. of which a = 25 and r = 4. 

17. The 20th term of the G. P. of which a = 1 and r = 1.04. 

18. The 30th term of the G. P. of which a = 50 and r = 1.02. 

269. The terms of a geometric progression between any two 
other terms are called the geometric means of those two terms. 

Thus, the three geometric means of 2 and 162 are 6, 18, and 54, 
since 2, 6, 18, 54, 162 form a geometric progression. 





GEOMETRIC PROGRESSION 


391 


A single geometric mean of two numbers is particularly im¬ 
portant. It is called the geometric mean of the numbers. 

When two numbers are given, any specified number of geo¬ 
metric means may be inserted between them. 

Example. — Insert three real geometric means between 9 
and ^. 

Solution. — 1. There results a geometric progression of 5 terms, in 
which a = 9, l = and n = 5. Find r. 

2. I = ar n_1 . .*. =_9 • r 4 , or r 4 = . 

r = = ^ f • 

3. The progression is: 9, 6, 4, or 9, — 6, 4, — |, 

Note. — The other two fourth roots of are imaginary, and imagi¬ 
nary means result from using them. 

Check. — Each G. P. has three terms between 9 and / 

EXERCISE 200 

1. Insert 4 geometric means between 1 and 243. 

2. Insert 5 geometric means between 1 and 64. 

3. Insert 2 geometric means between and 4. 

4. Find the geometric mean of 9 and 81. 

5. Find the geometric mean of 4 y and 16 y 1 . 

6. Find the geometric mean of 4 n and n. 

' . . a ■ b 2 

7. Find the geometric mean of — and — . 

b l a 

8. Find the geometric mean of x and y. 

9. Insert 3 real geometric means between 4 and 16. 

10. Insert 2 real geometric means between x and y. 

11. Insert 3 real geometric means between a and b. 

12. Find by logarithms the ratio if 6 geometric means are 
inserted between 5 and 50. 

13. Find by logarithms the ratio if 4 geometric means are 
inserted between 15 and 900. 

14. Find by logarithms the ratio if 5 geometric means are 
inserted between 2 and 2000. 


K-y 

or 1 

h 




392 


ALGEBRA 


270. The sum of the first n terms of a geometric progression 

can be found without writing down the n terms. 

Given — the first term, a, the ratio, r, and the number of 
terms, n, of a geometric progression. 

Find — the sum of the n terms, S. 

Solution. — 1. S = a + ar + ar 2 • • • • + ar n ~ 2 + ar n ~ 1 

2. M r rS = ar + ar 2 •••• + ar n ~ l + ar n 

3. S — rS = a — — ar n 

4. (1 — r)S = a — ar n . 


5 . D (1 _ r) 



6. Since l = ar n ~ l , rl = ar n 



Example. — Find the sum of the first 6 terms of 2, 6, 18, •••• 
Solution. — 1. a = 2, r = 3, n = 6. Find S. 

? q _ « — ar n . s _ 2 - 2 • 3 6 _ 2 - 1458 _ - 1456 _ ?0 o 
1 — 3 1 — 3 —2 —2 


EXERCISE 201 
Find the sum of the first: 

1. Seven terms of the progression 6, 12, 24, •••• 

2. Eight terms of the progression 20, 10, 5, — 

3. Six terms of the progression 4, — 12, +36, — • 

4. Five terms of the progression — T ^-, •••• 

6. Five terms of the progression — 4, 20, — 100, •••• 

6. Twelve terms of the progression 1, x 2 , x 4 , a: 6 , •••• 

7. Fourteen terms of the progression 5, 5 b 2 , 5 b 4 , •••• 

8. Find the sum of 20 terms of 1, (1 + r), (1 + r) 2 , •••• 

9. Find the sum of the first 10 powers of 2. 










GEOMETRIC PROGRESSION 


393 


10. Find the sum of the first 10 powers of 3. 

11 . a. Find the sum of 10 terms of 1, 1.04, (1.04) 2 , (1.04) 3 , ••••. 
b. If you have had the chapter on logarithms, find the value 

of this sum to three decimal places. 

12. a. Find the sum of 25 terms of 1, 1.025, (1.025) 2 , (1.025) 3 , 


b. By logarithms find the value of this sum. 

13. What is the sum of the 7th to the 12th terms, inclusive, 
of the geometric progression whose first term is 15 and whose 
ratio is V2? 


14. If a man were to deposit in a savings bank $5 on the 2d 
of January, $10 on the 2d of February, $20 on the 2d of March, 
how much would he have deposited by the end of the year? 


A" 

fa* 


y) 


15. Suppose a boy were to save 2^ during the year when he 
is five years old, 4^ when he is six years old, 8^ when he is seven 
years old, etc. How much would he have when he became 22 
years old ? 

SUPPLEMENTARY TOPIC ^ IV 

271. An infinite geometric progression is one the number of 
whose terms is infinite (uncountable). 

If the ratio of the progression is greater than 1 (for example, 
1.5), then the terms become larger and larger, and the sum of 
an infinite number of the terms is infinitely large. 

If the ratio of the progression is 1, then every term equals 
the first term. Again, however, the sum of an infinite number 
of terms will be infinitely large. 


Thus, consider the progression 3, 3, 3, 3, —. In it, the ratio is 1. 
The sum of a million terms is 3,000,000. The sum of an infinite num¬ 
ber of terms is 3 times an infinitely large number; therefore, the sum 
is infinitely large. 


When the ratio is less than 1 , however, the progression is 
especially interesting. 


394 


ALGEBRA 


Example 1 . — Consider the progression 5, f, f, ••••. What 
happens as the number of terms increases infinitely? 

Solution. — 1 . The ratio is 

2. When n = 4, l = ar n_1 = 5 (£) 3 = 

c _ a — rl _ 5 — ^ • -gj _ 5 — 

" 1 - r 1 -i 1 -i 

3. When n = 10, 1 = 5(i ) 9 = —^— 

19,683 

5 _ i . 5 

S = a ~ ^ _ 3 19,683 

n 1 -r “ 1 - i 

5 - —— 

= 59,049 

1 — i 


4. As n increases, l decreases; also the term rl of S decreases. 

5. If n were to become infinitely large, l would become approxi¬ 
mately zero; therefore rl also would become approximately zero. 


6 . 


/. S n is almost 



when n is infinitely large. 


Consider now the geometric progression in which the first 
term is a and the ratio, r, is less than 1. 

The sum of n terms is S n = a ~ ar • 

1 — r 

Since r is less than 1, r n decreases as n increases and becomes 
approximately zero as n becomes infinitely large. 


For example, the 30th power of £ is J q 73 y 41 824 and this is certainly 

almost zero. Moreover, remember this is only the 30th power of 
How much smaller even the 60th power would be! And an infinitely 
large power, — how negligibly small it would be! 


No matter what a may be, if it is multiplied by a number 
which is approximately zero, the product also is approximately 
zero. Therefore ar n is approximately zero, when n is infinitely 
large. 

S n becomes approximately - a - ~ Q or —-— 

1 — r 1 — r 




















GEOMETRIC PROGRESSION 


395 


Hence the sum of an infinite number of terms of a geometric 
progression in which r is numerically less than 1 is given by the 
formula 

S = —-—• 


1 - r 

Example 2. — What is the sum of an infinite number of terms 
of 4 — s is ....? 

Solution. — 1 . - f + 4 = - also (^) -*-(-§) = - f. 


r 


2. .*. the absolute value of r is f, which is less than 1. a = 4 


/. S = 


4 = 4 = ^ op24 

1 H - ■§■ ^ 5 


Example 3. — What is the sum of an infinite number of terms 
of the progression 1 , -J-, -J-, ••••? 

Solution. — 1. a = 1; r = £; S = —-— 


1 — r 


2 . 


S = 


1 - 


= - = 2 . 


Check. — Does l+£ + i + i + 1 3 s + T* + -" approximately equal 2 ? 
Yes, as you will easily see if you think of these numbers as 1", 
i", etc. 

EXERCISE 202 


Find the sum to infinity of: 
1 - 2 , •••• 

2* 3, 1, -j, •••• 

3. 12, 3, f, •••• 

4. 5, .5, .05, •••• 

fill i .... 

nr> 


6 . y, K y ~, •••• 

3 9 

_ 2 m2 W? 

7 . m , —,-, 

10 100 


3 ’ 


V’ 


4, — — Yg-, 

10 - b ~ **** 

11 . Find the sum of the repeating decimal .8181* 

81 


Solution. 


8181 100 + 10,000 


+ etc. 


81 


2. This is a geometric progression in which a = and r = 








396 


ALGEBRA 


The value of the decimal if an infinite number of places is considered 
is given then by the formula 



3. The value is 


' 81 

100 _ 81 .99 

_ 1 _ 100 ' 100 

100 


81 

99 


9_ 
11 * 


Find the value of the following repeating decimals: 

12. .6666-- 15. .8333*•• 18. .1111 — 

13. .4444— 16. .409090-- 19. .24333— 

14. .636363-- 17. .3555-- 20. .52222- 









XXIII. THE BINOMIAL THEOREM 

272. The binomial theorem is a formula for expanding any 
power of a binomial. 

By actual multiplication: 

(a + x) 2 = a 2 + 2 ax + z 2 . ( 1 ) 

(a + x ) 3 = a 3 + 3 a 2 x + 3 ax 2 + x 3 . (2) 

(a -f x) 4 = a 4 + 4 a 3 x + 6 a 2 x 2 + 4 ax 3 + x 4 . (3) 

Rule. — To expand any power of a binomial, like (a -f- x) n : 

1. The exponent of a in the first term is n and decreases by 1 
in each succeeding term until it becomes 1 in the next to the 
last term. 

2. The first term does not contain x. The exponent of x in 
the second term is 1 and increases by 1 in each succeeding term 
until it becomes n in the last term. 

3. The coefficient of the first term is 1; of the second is n. 

4. If the coefficient of any term be multiplied by the exponent 
of a in that term, and the product be divided by the number of 
the term, the quotient is the coefficient of the next term. 

Example 1 . — Expand (a + x) b . 

Solution. — 1. The exponents of a are 5, 4, 3, 2, 1. The exponents 
of x, starting with 1 in the second term, are 1 , 2, 3, 4, and 5. Writing 
the terms without the coefficients gives: 

a 5 + (?)a 4 x + (?)a 3 x i + (?)aV + ( ?)ax 4 + x 5 . 

2 . The 1 ’coefficient of the first term is 1 , and of the second term is 5 
(Rule 3). Multiplying 5, the coefficient of the second term, by 4, the 
exponent of a in the second term, and dividing by 2 , the number of the 
term, gives 10 , the coefficient of the third term; and so on. 

Filling in the coefficients in this manner gives: 

(o + x ) 5 = a 5 + 5 a 4 x + 10 a?x 2 + 10 a?x? + 5 ax 4 + x 5 . 

397 


398 


ALGEBRA 


Observe that the number of terms is n + 1. 

Thus, there are 6 terms in Example 1, in which n = 5. 

Observe also that the coefficients of terms “ equidistant ” 
from the ends are the same; for example, the second and the 
next to the last, etc. 


/ m\6 

Example 2. — Expand (2 — —J 
Solution. — 1. In this example, a is 2 and x is (-?> 

2 . ( 2 - f )' = 26 + 6 7 8 9 10 ' 25 ' (“ f ) + 16 ' 24 ' (“ f ) 1 

<-?) 


3. = 64 + 6 • 32 


4. = 64 — 64 m + 


+ 15 • 16 ■ — + 20 • 8 


(-») 




+ 12 ( 


m a \ 

243/ + 729 


80 


160 3 , 

27 “ + 


3 , 20 4 
27 m 


4 w 6 

81 W + 729* 


Note 1. — When the second term of the binomial is negative, the * 
terms of the expansion are alternately positive and negative. 

Note 2. — When the terms of the binomial are complicated mo¬ 
nomials, place each in parentheses, and afterwards simplify as in Steps 
3 and 4. 


EXERCISE 203 

Expand the following: 


1. (m + w) 5 

2 . (x - y ) 4 

3. (x + l ) 5 

4. (b - 2 ) 4 

5. (r 2 - « 2 ) 5 


6 . (a - 2 by 

7. (2 6 + c ) 5 

8 . (1 + y) s 

9. (1 - y 2 ) 6 

10. (a - 4 6) 4 


11. (» - i ) 4 

12 . (* + a 2 ) 6 

13. (2 x - 3 ) 5 

14. (ox 2 + 6) 5 

15. (4 + x 3 ) 4 


THE BINOMIAL THEOREM 


399 


Find the first three terms of: 


16 . (m - 4 ) 10 

20 . (a + 3 ft 4 ) 16 

23 - fS 


17 . (x 2 + 2 y) n 

21 . (a 2 - 2 6 3 ) 8 

\2 

3/ 

18 . (o - £) 20 

/x , y \ 8 

_ . /I 

\ 6 

19 . (m — n 2 ) 10 


24 . {- 

-y) 


25. Find the approximate value of (1.05) 10 to four decimal 
places. 

Solution. — f. (1.05) 10 = (1 + .05) 10 

2. (1 + .05) 10 = l 10 + 10-1 9 • .05 + 45*l 8 • (.05) 2 + 120-1 7 • (.05) 3 

+ 210-1 6 • (.05) 4 

3. = 1 + 10-1 • .05 + 45-1 • .0025 + 120-1 • .000125 

+ 210-1 • .00000625+•••■ 

4. = 1 + .50 + .1125 + .0150 + .0013125 

5. = 1.6288125, or 1.6288 

Find to four decimal places : 

26. (1.01) 10 27. (1.03) 15 28. (1.04) 10 29. (1.02) 20 

30. Write the first four terms of (a — b) n . 


SUPPLEMENTARY TOPICS 


273. The rth or general term of (a + x) n . Following the 
rules of § 272, 

■ . i n(n — 1) « « 

(a + x) n = a n + n • a n ~ l x + ——- — • a n 2 ar 


+ 


~ ' I arV + 


Note the fourth term. The exponent of £ is 1 less than the 
number of the term; the exponent of a is n minus the exponent 
of x ; the last factor of the denominator equals the exponent of 
x ; in the numerator there are as many factors as there are 
factors in the denominator. The rule follows. 




400 


ALGEBRA 


Rule. — In the rth term of (a + x ) n : 

1 . The exponent of x is r — 1 . 

2. The exponent of a is n — the exponent of x,i.e.,n — r+ 1. 

3. The denominator of the coefficient is 1 • 2 • 3 ••• ( r— l), 
the last factor being the same as the exponent of x. 

4. The numerator of the coefficient is n(n — l) ••• etc., until 
there are as many factors as in the denominator. 

n(n 1) 


The rth term is 

Example. 


^ H~ 2) # r+l . x r ~ l . 


1 • 2 ••• (r - 1) 

— Find the 8 th term of (3 a^ — b) 11 . 

Solution . — 1 . (3 a? — b) u = {(3 a*) + (- &)} 11 
2. In the 8 th term, the exponent of (—6) will be 7 (Rule 1); the ex¬ 
ponent of (3 a will be 11 — 7, or 4; the last factor of the denominator 
will be 7, and there will be 7 factors in the numerator starting with 
11 - 10 , etc. 


3. 


The 8 th term is 11 ‘ *° ‘ 9 'f ' 7 ' 0 J 5 • (3 a*) 4 (- b ) 7 


1-2-3-4-0-0-7 
or 330(81 d 2 )(- b 7 ) = - 26730 a 2 b 7 . 

Note. — If the second term of the binomial is negative, it should be 
inclosed, sign and all, in parentheses, before applying the rules. Also, 
if either term has an exponent or coefficient other than 1 , the term 
should be inclosed in parentheses before applying the rules. 

EXERCISE 204 


A' 


Find the 

. 



1 . 

5th 

term 

of 

(m + ») 10 

8 . 

2 . 

7th 

term 

of 

(a - 6) 9 

9. 

3. 

10 th term of (m — n) 15 

10 . 

4. 

4th 

term 

of 

(d + 3 ) 8 

6 . 

9th 

term 

of 

(b 3 + a 2 ) 14 


6 . 

8 th 

term 

of 

(2 x 2 - yT 

11 . 

7. 

7 th 

term 

of 

(3x - 2) 10 

12 . 


\ 


1\1Z 

•27 


l term of (- — a) 
v a / 

1 term of d ~ p) 

12 . 7th term of (d? — 2 c 2 ) 11 


2\ 10 




THE BINOMIAL THEOREM 


401 


274. The binomial formula has not been proved in this 
chapter; it has been written down from observation of the 
results in certain special cases. The formula has been applied 
only for positive integral values of n. 

The proof of the formula for positive integral exponents will 
be found in § 276. 

In more advanced courses in mathematics, the formula is 
proved to be correct (with certain limitations) not only for 
positive integral values of n but also for negative and fractional 
values. 

Historical Note. — The binomial theorem was formulated by 
Newton. 

275. Compound interest. When interest on a sum of money 
invested for one interest period is added to the principal, and, 
with it, draws interest during the next interest period, and so 
on, then the money is said to be invested at compound interest. 

Thus, the interest on $1 at 4 % for 1 year is $.04. If this is added to 
$1, making $1.04, and if $1.04 draws interest for one year, the interest 
during the second year is $.0416, and the amount at the end of two 
years is $1.04 + .0416 or $1.0816. 

This is called the compound amount, and $.0816 is called the com¬ 
pound interest. The interest was compounded annually at 4%. 

Problem. — Find the compound amount at the end of n years 
if 1 dollar is invested at rate r compounded annually. 

Note. — “ Rate r ” means 4%, or 5%, or 3£%, etc. 

Solution. — 1. The interest during the first year is r, and the amount 
at the end of year 1 is 1 + r. 

2. The interest for the second year is r(l + r). 

.*. the amount at the end of year two is (l+r)+r(l+r), or 

(1 + r)\ 

3. The interest for the third year is r(l + r) 2 . 

/. the amount at the end of the third year is (1 + r) 2 + r(l + r) 2 . 

This equals (1 + r)(l + r) 2 , or (1 + r) 3 . 

4. Similarly, the compound amount at the end of 4 years is (1 + r) 4 ; 
at the end of 5 years is (1 + r) 5 ; and at the end of n years is (1 -f- r) n . 


402 


ALGEBRA 


Example. — What is the compound amount of $250 invested 
at 4 % compounded annually for 10 years ? 

Solution. — 1. In this example, r = .04 and n = 10. 

2. .'. the compound amount of $1 = (1 + .04) 10 . 

3. You have found the value of this expression to four decimal places 
in Example 28, page 399. It is $1.4802. 

4. the compound amount of $250 at 4% compounded annually 
for 10 years is 250 X 1.4802, or $370.05. 

Note. — The value of (1.04) 10 can be obtained either by the binomial 
theorem or by logarithms. 


EXERCISE 205 

Find the compound amount of: 

1 . $500 at 4% compounded annually for 10 years. 

2 . $500 at 2 % compounded annually for 20 years. 

3 . $1000 at 6 % compounded annually for 5 years. 

4 . $1000 at 3 % compounded annually for 10 years. 

Note. — This section gives a mere introduction to the application 
of the binomial theorem to one of the most important business applica¬ 
tions of algebra. 

The mathematics of investment and actuarial mathematics are founded 
on the mathematics taught in our chapters on exponents, logarithms, 
progressions (especially geometric), and the binomial theorem. Any 
student contemplating a course in commerce in any university will 
find knowledge of these chapters indispensable. 


SUPPLEMENTARY TOPICS 


276. Proof of the binomial theorem for positive integral 
exponents. Assume, as in § 272, that 

(a + x) n = a n + na n ~ l x + -— a n ~ 2 x 2 

1*2 

n(»-l)(n-2) ^ (1) 

1 •2 • 3 3 w 






THE BINOMIAL THEOREM 


403 


Multiply both members of equation ( 1 ) by a + x. Then 
(a + x) n+1 = a n+1 + na n x + ^ ~ * l \ n ~ l x 2 


1 -2 


+ 


n(n - 1 ){n - 2 ) „ n _ ; 


1-2*3 

n — 
~2 


a n ~ 2 * x z + 


a"x + ja n ~V + ^L—Har-V + 


(a + x ) n+1 = o " +1 + (n + 1 )a n x + n | ^ + 1 1 a” _I x 2 

, n(n — \) \ n — 2 . .,1 „ 2 •> 

+ -W{— +1 |° x 

= a” +I + (n + l)a n x + n • —i—L^x 2 

I n(n - 1 ) . n + 1 „_ 2 3 . ... 

+ 1-2 “ I - + 

= a " +1 + (ra + l)a“x + (” + !)' re a n-i^ 

1*2 


(w + 1) - w (n - 1) „ n __; 


1-2-3 


a n ~ 2 ot? + 


Observe that the expansion on the right is in accordance with 
the rules of § 272. This proves that, if the rules are assumed 
for any particular positive integer, n, they hold true also for 
the next greater integer, n + 1 . 

But the rules are known to be satisfactory in the case of 
(a + x ) A ; hence they hold for (a + z) 5 . Since they hold for 
(a + x) 5 , then they hold also for (a -f x) 6 ; and so on. 

Therefore the binomial theorem is true for any positive 
integer. 


277. Fractional and negative exponents. In the expan¬ 
sion of § 272, if n is a positive integer, there is ultimately 
a term (the (n -f 2 )nd), for and after which the coefficient 
n(n — l)(n — 2 ) 


is zero. 

















404 


ALGEBRA 


When n is a negative integer or a fraction, there is no term 
for which the coefficient is zero. Hence the terms continue 
indefinitely. The resulting expansion has an infinite number 
of terms. 

In this case also, the expansion on the right in § 272 has 
a sum, and this sum is ( a + x) n for any rational (§ 214) value 
of n, provided the absolute value of a is greater than the abso¬ 
lute value of x. This theorem is proved in a more advanced 
course in mathematics. 

Assuming the theorem, the following examples may be solved : 

Example 1 . — Expand ( a -f to four terms. 


Solution. — 1 . Substitute § for n in the formula. 

2. /. (a + x)i- 

,l)(f - 2 ) qf-3^3 


= a* + -av-'x + t(f - + 


1 • 2 


1-2-3 


3. = af +-aT*x — ^aT^x* t ?. g-itf -|- 

3 2 1 • 2 • 3 

4. = a i + |a-*x-ia-tr= + ia-^ + ... 


Example 2. — Find the 7th term of (a — 3 x~*) 

Solution. — 1. The 7th term may be found by applying the rule in 
§273. 

2. Substitute (— 3 x~%) for x, and (— £) for n. 

The exponent of (— 3 aT$) is 7 — 1 or 6. 

The exponent of a is — £ — 6 or — 

The denominator of the coefficient isl-2-3-4-5-6. 

The numerator of the coefficient is (— £) ( — £ — 1) • • • until there are 
six factors. 


Hence the seventh term is: 

$)(~ J srH~ •¥•)("" (_ 


• 2 • 3 • 4 • 5 

f--*• 

_ 728 
9 


• 6 

3 6 x “ 9 


(o"^)(-3a;“t) # 


— ——a ~^~x' 







THE BINOMIAL THEOREM 


405 


EXERCISE 206 

Find the first four terms of: 

1. (a + x )* 4. 'V / a — b 

2. (1 + x)~* 5. (a* + 2 b)$ 

3. (1 - x)~* 6. (a 3 - 4 x*)* 


7 ' 

(a + x ) 5 


Find the 

9. 6 th term of (a + x )* 

10 . 5th term of (a — 6 ) - * 

11 . 7th term of (1 + x )~ 7 

12 . 8 th term of (1 — x)* 


13. 9th term of (a — x)~ z 

14. 11th term of V(to + n ) 5 

15. 7th term of (a -2 — 2 6*)~ 2 

16. 8 th term of --- 

(x 6 + y-*y 


278. Extraction of roots. The binomial theorem may be 
used to find the approximate value of a root of a number. 


Example. — Find approximately to five decimal places. 
Solution. — 1. The nearest perfect cube to 25 is 27. 

2. ^25 = 'v / 27 - 2 = {(3 3 ) + (- 2)}* 

3. = (3 3 )i + i(3 3 )"t • (- 2) - i(3 Tk- 2 ) 2 

+ ^r(3 3 )~^(— 2) 3 + ••• 


4. 

5. 


= 3 _2 4 40 

3 • 3 2 9 • 3 6 81 • 3 8 

= 3 - .07407 - .00183 - .00008, or 2.92402. 


Rule. — Separate the given number into two addends, the 
first of which is the least perfect power corresponding to the 
order of the root, and expand the result by the binomial theorem. 


EXERCISE 207 

Find the approximate values to four decimal places of: 

1 . Vl7 3. 6 . a/90 7. ^35 

2. VEl 4. \^75 6. ^275 8. ^60 













XXIV. VARIATION 


279. In § 139, a variable has been described as a changing 
number. 

Thus, the distance between a moving train and its destination varies , 
— that is, it decreases; the age of an individual varies from moment 
to moment, — that is, it increases. 

280. A quantity which is fixed in any given problem is called 
a constant. 

Thus, if a workman receives a fixed sum per day, the total wages due 
him changes from day to day if he works and remains unpaid. His 
daily wage is a constant; his total wages is a variable. 

281. A change or variation in one quantity usually produces 
a variation in one or more other quantities. Such variables 
are called related variables. For each value of one variable 
there is a corresponding value of the other variable, or variables. 

Thus, if the side of a square is increased, the perimeter and the area 
of the square are also increased. 

Instead of the quantities themselves, their measures in terms 
of certain units of measure are used. 

Thus, distance is expressed as a number of miles, rods, or other units 
of length; weight is expressed as a number of units of weight; area is 
expressed as a number of units of area. 

282. One quantity varies directly as another when the 
quotient of any value of the one divided by the corresponding 
value of the other is constant. 

Thus, the ratio of the perimeter of a square to the side of the square 
is always 4, or the perimeter is 4 times the length of the side. If p 

406 


VARIATION 


407 


represents the perimeter and s represents the length of a side of the 

square, then - = 4, or p = 4 s expresses the fact that the perimeter 
s 

varies directly as the side of the square. 

Similarly, if x varies directly as y, then — = m or x = my, where m 
is a constant. ^ 

Thus, since C = ird in geometry, we may say, the circumference of a 
circle varies directly as the diameter. This means that the circumference 
is doubled if the diameter is doubled, or is multiplied by five if the 
diameter is multiplied by five. 

Again, y = mx 2 expresses the fact that y varies directly as the square 
of x. This means, that, if x is doubled, y is multiplied by four; if x is 
multiplied by 3, y is multiplied by 9; etc. 

283. One quantity is said to vary inversely as another when 
the product of any value of the one and the corresponding value 
of the other is constant. 

Thus, the time and rate of a train going a distance d are connected 
by the equation rt = d. If the distance remains fixed, then the time 
varies inversely as the rate; for example, if the rate is doubled, the 
time is halved. 

If x varies inversely as y , then xy = m, where m is a con¬ 
stant, expresses the relation between them. 

284. One quantity is said to vary jointly as two others when 
it varies directly as their product. If x varies jointly as y and 

z, then — = m, where m is a constant, expresses the relation 

\ 

between the variables. 

Thus, the wages of a workman varies jointly as the amount he re¬ 
ceives per day and the number of days he works; for, letting W equal 
his total wages, w his daily pay, and n the number of days he works, 
then W = nw. Here m = 1. 

Again, the formula for the area of a triangle is 
A = % ab. 

This shows that the area of a triangle varies jointly as the base and 
altitude. (Here m = £.) 


408 


ALGEBRA 


285. Expressing more complicated relations between 
variables. 

x varies directly as y and inversely as z is expressed thus: 
— = m or xz = my, where m is a constant. 

y 

x z varies inversely as the square of y is expressed thus: x?y 2 
= m, where m is a constant. 

EXERCISE 208 

Express in words the following relations, using the expres¬ 
sions “ varies directly/’ “ varies inversely,” and “ varies 
jointly.” 

1. A = ab, where A = the area, a = the altitude, and b = 
the base of a rectangle, and where A, a, and b are regarded as 
variables. 

2 . If in Example 1 , A and b are variable, and a is constant, 
express the relation between A and b. 

3. If, in Example 1 , A is constant, but a and b are variables, 
express the relation between a and b. 

4. C = 7 Tr 2 , where C and r are variable and tt = 3.1416. 

5. V = Iwh, where V, l, w, and h are all variable. 

6 . V = Iwh, where l is constant, but V, w, and h are variable. 

7. V = Iwh, where l and w are constant, but V and h are 
variable. 

8 . I = PRT, where I, P, R, and T are all variable. 

9. a. I = PRT, where P and R are constant, but I and T 
are variable. 

b. In this case, if T is trebled, what happens to I ? 

10 . a. I = PR T, where P is constant, but I, R, and T are 
variable. 

6 . In this case, suppose R is doubled and T is trebled, what 
happens to /? 


VARIATION 


409 


11 . a. V = -J 7rr 3 , where V = the volume and r = the radius 
of a sphere, and V and r are considered variable. 

b. In this case, if r is doubled, what happens to F? 

12. a. nc = C, where n = a number of articles, c = the cost 
of each, and C = the total cost, where C is assumed as con¬ 
stant, but n and c are considered variable. 

b. In this case, what happens to c if n is doubled ? 

c. What happens to n if c is doubled ? 

13. a. If, in Example 12, n is constant, but c and C are 
variable, express the relation. 

b. In this case, if C is multiplied by 5, what happens to c ? 

c. If c is multiplied by 4, what happens to C ? 

14. a. S = £ CL, where S = the lateral area, C = the cir¬ 
cumference of the base, and L = the slant height of a right 
circular cone, where S, C, and L are considered variable. 

b. In this example, if C and L are each doubled, what happens 
to S? 

15. The per capita cost of instruction of pupils in a room 
varies directly as the salary of the teacher, and inversely as 
the number of the pupils. Select letters to represent the 
variables, and express this relation by an equation. 

16. a. The volume of a cone varies jointly as the altitude 
and the area of the base. Select letters to represent the vari¬ 
ables, and express this relation by an equation. (This equation 
will have in it one letter representing a constant.) 

b. Assuming the truth of the equation you have written in 
part a, find the value of the constant if you are told that a cone 
of base 75, and altitude 5, has volume 125. 

17. a. The distance a falling body falls from a position of 
rest varies as the square of the number of seconds it falls. Using 
suitable letters to represent the variables, express this relation 
by an equation. 




410 


ALGEBRA 


b. If you are told that the distance is 513.28 ft. when the time 
is 4 seconds, what is the value of the constant you have used in 
part a? 

c. Using the constant you have discovered in part b, find the 
distance when the time is 5 seconds. 

18. y varies directly as x, and is 40 when x is 4. What is the 
value of y when x is 9 ? 

19. y varies directly as x 2 and is equal to 60 when z is 3. 
What is y when x is 5 ? 

20. x varies inversely as y, and is f when y is f. What is the 
value of x when y is § ? 




XXV. DETERMINANTS 


286. The symbol 


3 4 
2 7 


is called a determinant. Its value 


is defined to be 3 • 7 —2*4, which equals 21 — 8, or 13. 
a c 
b d t 

' a c 
b d 


In general 
and is defined thus : 


is called a determinant of the second order 

= ad — be. 


The numbers a, b, c, and d are called the elements of the 
determinant. 

Clearly, any difference such as rs — mn may be arranged as 

r n 
t m s 

2 -51 


a determinant: thus rs 
Example 1. 


,4 +3 . 

Example 2. — 26 - 15 = 2 • 13 


2 • 3 — 4(— 5) = 6 + 20 = 26. 
3 • 5 = 


2 5 

3 13 


EXERCISE 209 


2. 


Find the values of: 

6 5 

4 2 , 

5 3 

2 - 7 


3. 

4 - 2 


5. 

-5 3 

7. 

2 m — p 

6 9 



2 6 


2 n r 

4. 

3 - ■ 

4 

6. 

3a 4 

8. 

3 a 4 a 1 

- 2 

7 

2c 1 


2 c 5 e 1 


Express as determinants: 

9. mn — xy 11. 33 — 14 13. cd + pq 

10. 2 ab — cd 12. 6 c — 5 d 14. 3 mn + 2 rs 

411 



























412 


ALGEBRA 


287. Solving simultaneous linear equations by means of 
determinants. 

(ax -f- by = c 
dx + ey = f 
Solved by elimination as in § 159, 

“ “ bf and y = af ~ cd 


Consider the system 


x = 


-bd 


ae — bd 



c b 


a c 

• q* — 

f e 


d f 


a b 

■ and y 

a b 


d e 


d e 


Notice that the two solutions may be expressed as the quo¬ 
tients of determinants whose terms are the coefficients of the 
equations. 

Rule. — To solve two simultaneous linear equations having 
two unknowns by determinants: 

1. Arrange the equations in the form : ] ax + &!/ = c - 

1 dx + ey = f. 

2. The value of x is a fraction: its denominator is the deter¬ 
minant formed by the coefficients of x and y, a b ; its nu- 

d e 

merator is the determinant obtained by replacing the coefficients 
of x in the denominator determinant by the corresponding abso¬ 
lute terms, c ^ 
f e 

3. The value of y is a fraction with the same denominator as 
x ; its numerator is the determinant obtained by replacing the 
coefficients of y in the denominator determinant by the absolute 

a c 


terms, 


d f 


Example. — Solve by determinants the pair of equations: 

2 x — by = - 16. 

3 z + 7 y = 5. 
















DETERMINANTS 


413 


Solution. — 1 . 


- 16 - 5 I 

T - 5 7 I _ — 16 • 7 — 5( — 5) _ — 112 + 25 _ — 87 _ ? 

2-51 2-7 -3(-5) 14 + 15 29 

3 7 I 

2 . 

12 - 16 

J_3_5 _ 10 - 3( — 16) _ 10 + 48 _ 58 _ 0 

V 12 - 5 2-7 - 3(- 5) 14 + 15 29 

I 3 7 

Check. — In (1) : Does 2(- 3) - 5(2) = - 16? Does - 6 - 10 = 
-16? Yes. 

In (2): Does 3(- 3) + 7(2) = 5? Does - 9 + 14 = 5? Yes. 


EXERCISE 210 

1 - 10 . Solve by means of determinants Examples 1-10 of 
Exercise 109. Check the solutions. 

288. Determinants are especially useful in solving simul¬ 
taneous linear equations with more than two unknowns. 


ai 

02 

03 

h 

b 2 

bz 

Cl 

c 2 

Cz 


is called a determinant of the third order. Its value is defined 
to be: 

«1&2 C 3 4 " ®2^3 C 1 4 " O3C2&1 — £162^3 “ ^l a 2 c 3 — ®1^3 C 2 

The adjoining diagram aids in recalling this 
value. Take the product ai& 2 C 3 along the di¬ 
agonal and add to it the two products formed 
by starting with a 2 and a 3 respectively and fol¬ 
lowing the arrows which point in the direction 
of this diagonal; then subtract the product 
cibidz along the other diagonal, and also sub¬ 
tract the two other products formed by starting 
with 61 and a\ respectively and following the 
arrows which point in the direction of this sec¬ 
ond diagonal. 


















414 


ALGEBRA 


Example. — 


1 5 2 
4 7 3 
2-36 


= 1 • 7 • 6 + 5 - 3 - 2 + 2(- 3) - 4 
-2-7-2 -4*5-6 
- 1 - 3-(-3) 

= 42 + 30 - 24 - 28 - 120 +9 
= - 91. 


EXERCISE 211 

Find the value of: 


1 2 3 


2 4 6 


2 2 3 

2 1 2 

2. 

3-23 

3. 

-2 -4 - 11 

3 3 1 


1 5 4 


5-6 2 


4. Solve by determinants the equations : 

x + y — z = 14 
jz + 3?/— z = 16 
\x + y — 3 J 2 = — 10 


Solution. — A rule similar to that of § 287 applies for linear equations 
with more than two unknowns. Namely: 


14 1 -1 
16 3-1 
- 10 1-3 

- 126 + 10 - 16 -30 + 48 + 14 - 100 

3 1-1 

-27-1-1+3+3+3 -20 

13-1 


1 1 -3 



3 

14 

- 1 



3 

1 

14 


1 

16 

1 



1 

3 

16 


1 

- 10 

-3 

_ -i20_ 6 

- 

1 

1 

- 10 

- 140 

3 

1 

- 1 

-20 b - 

<6 - 

3 

1 

- 1 

-20 

1 

3 

- 1 



1 

3 

- 1 


1 

1 

-3 



1 

1 

-3 



Check. — The solution checks when substituted in the three equations. 
Note. — The equations must be arranged first in the form ax + by 
+ cz = d. Thus the equation 2 x — 3 z = 7 would be written 2 x + 
0 y — 3 z = 7. 
























DETERMINANTS 


415 


Solve the following equations by determinants: 


[ x + y - z = 24 

5. <4o; + 32/ — s = 61 
[Gx—5y—z = ll 
\5x - i/+4z= —5 

6 . 3z+5i/ + 6 z = -20 

[ x —|— 3 2 / — 82 = 27 

14 a — 5 6 — 6 c = 22 

7 . <a — b + c = —6 
I9a + c =22 


[4 x — 3 y = 1 
8 . 1 4 y — 3 z = — 15 
[4 z — 3 £ = 10 
f 9 £ + 5 z = — 7 
9C < 3 £ + 5 y = 1 
19 y + 3 z = 2 
\2x + 5y+3z = — 7 
10. <2y—4z=2—3x 
[5 a; H- 9 2 / = 5 + 7z 


XXVI. 


TRIGONOMETRY OF THE RIGHT TRIANGLE 


This subject is included as part of the unit in elementary algebra 
recently proposed by the College Entrance Requirements Board, and by 
the National Committee on Mathematical Requirements. The fol¬ 
lowing material is designed for those who wish to comply with this re¬ 
quirement. It can be studied at any time after the early work in 
fractions has been studied, — preferably after the sections devoted to 
ratio and proportion. 



289. Sine of an angle. Carefully make 
a tracing of the angle ABC of the figure at 
the right. Extend the sides BA and BC 
until they are about 5 inches long. (Keep 
this figure.) 

Take point P\ so that BPi is 3 inches. Draw P\R\ perpen¬ 
dicular (at right angles) to BA. Measure P\Ri to sixteenths of 
P R 

an inch. Find the ratio to two decimal places by divid- 
BPi 

ing the length of PiRi by that of BP\. 

Now take P 2 so that BP 2 is 4 inches. Draw P 2 R 2 perpen¬ 
dicular to BA. Measure P 2 R 2 to sixteenths of an inch. Find 
P 2 R 2 


the ratio 


PoB 


Compare the values you have found for these two ratios. 
They should be equal at least to one decimal place. 

Wherever you place a point P on line BC, when you draw the 
perpendicular PR, the ratio of it to the distance BP will always 
have this same value. This ratio is called the Sine of angle B. 
(Sin B). In the right triangle formed by lines BR, BP, and PR, 
PR (the side opposite angle B) 


sin B = 


BP (the hypotenuse of the triangle) 

416 







TRIGONOMETRY OF THE RIGHT TRIANGLE 417 


The value of this ratio changes as the angle changes. The 
values of the sine for every degree of angle from 1° to 89° in¬ 
clusive are in the columns headed Sin of the table on page 424. 
Notice that the sine increases as the angle increases. 

Thus sin 63° = .891. 

This means that the side opposite a 63° angle of a right triangle, 
divided by the hypotenuse of the triangle, equals .891. 

Again, if sin x = .682, then the angle must be 43°. 

The approximate value of the sine of any angle between 1° 
and 89° can be found by use of this table. 

Example. — Find sin 27° 25'. 

Solution. — 1. sin 27° = .454 ) 

sin 27° 25' = ? > difference = .015 

sin 28° = .469 J 

2. 25' = U or & of 1°. ft X .015 = .006 

3. Since the sine increases as the angle increases, 

sin 27° 25' = .454 + .006 = .460 

Note. — This process is called interpolation. 

290. Cosine of an angle. In the figure you drew for § 289 
measure BRi and also BR%. 

T> T> 

Find: a. the ratio —to two decimals. 

BP\ 

B /? 

b. the ratio to two decimals. 

B P 2 

Your results, correct to one decimal place, should be the same 
for the two ratios. 

Wherever P is taken on BC, the ratio of BR to BP will have 
the same value. This ratio is called the cosine of angle B 
(Cos B). In the right triangle BPR, BR is the side adjacent to 
angle B, and BP is the hypotenuse. 

In a right triangle BPR , 

Cos B — BR s ^ e adjacent to angle B) . 

BP {the hypotenuse of the triangle) 



418 


ALGEBRA 


The value of the cosine of an angle changes as the angle 
changes. These values are found in the columns headed Cos on 
page 424 for every degree of angle from 1° to 89°. 

Thus, cos 36° = .809. 

Notice that the cosine decreases as the angle increases. 

The approximate value of the cosine of any angle between 
1° and 89° can be found by use of this table. 

Example. — Find Cos 62° 15'. 

Solution. — 1. Cos 62° = .469 1 

Cos 62° 15' = ? [ difference = .015 

Cos 63° = .454 J 

2. 15' = | of a degree. 1 of .015 = .0037 + , or .004 

3. Since the cosine decreases as the angle increases, 

cos 62° 15' = .469 - .004, or .465. 

291. Tangent of an angle. In the figure you drew for § 289, 
find : 

P R 

a. the ratio 11 to two decimal places. 

BRi 

P R*> 

b. the ratio to two decimal places. 

X>X2 

Your results, correct to one decimal place, should agree. 

Wherever P is taken, the ratio PR to BR will have the same 
value. This ratio is called the tangent of angle B (Tan B). 

In the right triangle BPR , 

tan B — PR (side opposite angle B) 

BR (side adjacent angle B) 

The value of this ratio changes as the angle changes. These 
values are found in the columns headed Tan on page 424 for 
every degree of angle from 1° to 89°. 

Thus, tan 27° = .510 

Notice that the tangent increases as the angle increases. 

The approximate value of the tangent of any angle between 
1° and 89° can be found by use of this table. 






TRIGONOMETRY OF THE RIGHT TRIANGLE 419 


Example. — Find tangent 72° 35'. 

Solution. — 1. Tan 72° = 3.078 ] 

Tan 72° 35' = ? [difference = .193 

Tan 73° = 3.271 J 

2. 35' = ££ or -jSg of a degree 

A x .193 = = .1125, or .113 

JLZ 

3. Since the tangent increases as the angle increases, 

Tan 72° 35' = 3.078 + .113 = 3.191. 


EXERCISE 212 

1. Obtain from the table: 

a. Sin 48°; sin 35°; 

b. Cos 34°; cos 67°; 

c. Tan 28°; tan 73°; 

2. What is the angle x : 

a. sin x = .438; sin x = .819; 

b. cos x = .946; cos x = .766; 

c. tan x = .344; tan x = .933; 


« 


sin 62°; 
cos 46°; 
tan 24°. 


sin x = .982 
cos x = .391 
tan x = 2.144. 


3. a. Find sin 35° 20'; 

b. Find sin 35° 40'; 

c. Find sin 35° 30'; 


d. sin 20° 15'; 

e. sin 76° 25'; 
/. sin 43° 50'. 


4. a. Find cos 40° 10'; 

b. Find cos 40° 20'; 

c. Find cos 40° 30'; 

5. Find a. tan 25° 30' 

6. Find a. sin 46° 15' 

7. Find a. cos 48° 20' 


d. cos 63° 25'; 

e. cos 15° 45'; 

/. cos 35° 5'. 

b. tan 25° 45'. 
b. sin 46° 35'. 
b. cos 48° 50'. 


8. a. What is the angle x when sin x = .635 ? 


Solution. — 1. 


sin 39° = .629 ] difference 
sin x = .635 / = .006 
sin 40° = .643 


I difference 
J = .014 


2. .006 .014 = .42+ 


.42 of 60' = 25.20' or 25' 
angle x = 39° + 25' (about). 


3. 



420 


ALGEBRA 


b. What is the angle x when sin x = .640 ? 

Find angle x, when: 

9. tan x = .350. 12. cos x = .880 

10. tana: = 2.165. 13. sin x = .925 

11. cos a: = .462. 14. tan x = 1.500 

Note. — If the value of the sine, cosine, or tangent of x were given, 

correct to four decimal places, a four place table would be needed to 
determine the angle x with a corresponding degree of accuracy. How¬ 
ever, the angle x can be determined with only slight error in such cases 
also, by using the three place table. 

Thus, if sin x = .3148, find x as if sin x = .315. 


292. Finding lengths, distances, and angles by means of the 
trigonometric ratios. 


Example 1. —Assume that AC represents a pole of unknown 
height; that B is 125 ft. from its foot; that the imaginary line 
AB makes with BC the angle measuring 34°. Find AC. 

Solution. — 1. Let AC = x feet. The ratio which 
uses the sides AC and BC is the tangent of angle B. 

Tan B = tan 34° = 

BC 125 b ^ 34 ' 

2. /. a: = 125 X tan 34° 

3. .’. x = 125 X .675, or 84.375 ft., or about 84 ft. 

Note. — Angle CBA is called the angle of elevation of point A at 

point B. 



Example 2. — AB represents a lighthouse 250 ft. high. AD 
is.an imaginary line parallel to BC. C represents the position of 
a ship. Angle DAC = 31°. Find BC. 


Solution. — 1. In such a figure, angle BCA 
equals angle DAC. (Make a tracing of angle DAC 
on tissue paper, and compare it with angle BCA.) 
:. angle BCA =31°. 



2 . 

3. 

4. 

5. D.601 


Tan BCA = — 
BC 

BC X tan 31°= 250 
BC X .601 = 250 
BC = 415. 


9 ft. 


/tan 31° = .601\ 
\in the table. / 







TRIGONOMETRY OF THE RIGHT TRIANGLE 421 


Note. — Angle DAC is called the angle of depression of point C at 
point A. 


Example 3. — In the adjoining figure, assume the pole to be 
40 ft. high and BC 30 ft. How long must the rope AB be made, 
and what angle does it make with BC ? 


Solution. — a. To find angle B. 

1. AC = 40, and BC = 30 

2. Tan B = .-. Tan B = = 1.333 

3. Tan 53° = 1.327 ) difference 

Tan B = 1.333 / = .006 

Tan 54° = 1.376 

•006 \ > _ 360 _ 

_X60 -^-7 + 

/. angle B = 53° 7' 


) difference 
= .049 



b. To find AB. 

1 . 


Sin B = — 


AC 

AB' 


2. .*. AB • Sin 53° 7' = 40 

3. /. .800 • AB = 40 

4. D. goo AB = 50. 


.*. Sin 53° 7'= — 
AB 


Note. — Sin 53° 7' = .800 is obtained as in Example 3, Exercise 212. 


EXERCISE 213 

1. In the adjoining figure, AC is perpendicu¬ 
lar to CD ; CD = 100 ft.; angle D = 62°. 

Find AC, and AD. 

2. 145 feet from the foot of a high building 
the angle of elevation of the top is 39°. How high is the 
building ? 

3. From the top of a hill known to be 150 ft. above the level 
of the plain, the angle of depression of a house is 22°. How 
far away is the house from an imaginary point directly below the 
top of the hill ? 







422 


ALGEBRA 


4. A boy is flying a kite at the end of 500 ft. of twine. 
Directly below the kite a second boy stands. The angle 
made by the twine (assumed to be stretched straight) with 
an imaginary line running from the first boy to the second is 
approximately 50°. Determine the approximate height of 
the kite. 


In triangle ABC, BC is 18 ft., angle B is 70°, and AB is 
Draw a triangle to represent these conditions and draw 


5. 

8 ft. 
the altitude AD. 

a. Find the length of AD. (Use the 
sine.) 

b. Having found AD, find the area. 

6. In the figure at the right is a 
right triangle in which AB = 10 ft. 
and BC = 15'. 

a. Find angle C. 

b. Find AC. 



7. At a point 175 ft. from the foot of a high building, the 
angle of elevation of the top is 51° 30'. How high is the build¬ 
ing? 


8. From a height of 350 ft., the angle of depression of an 
object on the plain below is 28° 15'. Find the distance of the 
object from a point in the plain directly below the point of 
observation. 


9. In triangle ABC, in which angle C is a right angle, and 
angle B is an angle of 24° 45', side AB is 24 in. long. Find the 
length of AC and of BC. 

10. The angle of elevation of an aeroplane at a certain point 
P is 38 30'. Point D, 1500 ft. distant, is directly below the 
aeroplane. How high is the aeroplane ? 





TRIGONOMETRY OF THE RIGHT TRIANGLE 423 

293. Logarithms simplify the computation. 

Example. — Find angle B when sin B = . 

494.7 

Solution. 1 . Find 265.3 -f- 494.7 by logarithms. 

log 265.3 = 2.4237 or 12.4237 - 10 
log 494.7 = 2.6943 2.6943 

log (265.3 494.7) = 9.7294 - 10 

265.3 -i- 494.7 = .5362, or .536 

2. .*. sin B = .536. Then angle B = 32° 24'. (page 419.) 

Note. — There are tables which give the logarithms of the sines, 
cosines, and tangents of angles. Their use simplifies the computation 
still more. Instruction in their use is more properly a part of a course 
in trigonometry. 

EXERCISE 214 

Use logarithms in the solution of the following examples: 


In Triangle ABC, in which Angle C = 90° 



BC 

AB 

AC 

Angle 5 

Angle A 

1 

846 

? 

625 

? 

? 

2 

975 

1154 

? 

? 

? 

3 

? 

683 

527 

? 

? 

4 

? 

? 

962 

36° 30' 

? 

5 

? 

378.5 

? 

42° 

? 

6 

265.8 

? 

425.3 

? 

? 

7 

483.7 

1292 

? 

? 

? 

8 

37.25 

68.43 

? 

? 

? 

9 

? 

29.75 

18.54 

? 

? 

10 

? 

361.7 

284.3 

? 

? 

11 

75 

? 

60 

? 

? 

12 

85 

100 

? 

? 

? 

13 

? 

90 

60 

• ? 

? • 

14 

? 

? 

120 

62° 

? 

15 

? 

150 

? 

36° 15' 

? 

16 

248 

? 

? 

? 

30° 

17 

? 

625 

? 

48° 30' 

? 

18 

56.5 

? 

? 

29° 45' 

? 




















1 ' 

2‘ 

3‘ 

4< 

5‘ 

6' 

r 

8' 

9’ 

10 ' 

11 

12 

13 

14' 

15 

16 

17' 

18 

19 

20 ' 

21 ‘ 

22 

23' 

24' 

25 

26 

27 

28 

29 

30 

31' 

32 

33 

34 

36 

36' 

37' 

38 

39 

40 

41 

42' 

43 

44' 

45' 


ALGEBRA 


Table of Values of Trigonometric Ratios 


Sin 

Cos 

Tan 

Angle 

Sin 

Cos 

Tan 

.018 

.999 

.018 

45° 

.707 

.707 

1.000 

.035 

.999 

.035 

46° 

.719 

.695 

1.036 

.052 

.998 

.052 

47° 

.731 

.682 

1.072 

.070 

.997 

.070 

48° 

.743 

.669 

1.111 

.087 

.996 

.087 

49° 

.755 

.656 

1.150 

.105 

.994 

.105 

50° 

.766 

.643 

1.192 

.122 

.992 

.123 

51° 

.777 

.629 

1.235 

.139 

.990 

.141 

52° 

.788 

.616 

1.280 

.156 

.988 

.158 

53° 

.799 

.602 

1.327 

.174 

.985 

.176 

54° 

.809 

.588 

1.376 

.191 

.982 

.194 

55° 

.819 

.574 

1.428 

.208 

.978 

.213 

56° 

.829 

.559 

1.483 

.225 

.974 

.231 

57° 

.839 

.545 

1.540 

.242 

.970 

.249 

58° 

.848 

.530 

1.600 

.259 

.966 

.268 

59° 

.857 

.515 

1.664 

.276 

.961 

.287 

60° 

.866 

.500 

1.732 

.292 

.956 

.306 

61° 

.875 

.485 

1.804 

.309 

.951 

.325 

62° 

.883 

.469 

1.881 

.326 

.946 

.344 

63° 

.891 

.454 

1.963 

.342 

.940 

.364 

64° 

.899 

.438 

2.050 

.358 

.934 

.384 

65° 

.906 

.423 

2.144 

.375 

.927 

.404 

66° 

.914 

.407 

2.246 

.391 

.921 

.424 

67° 

.921 

.391 

2.356 

.407 

.914 

.445 

68° 

.927 

.375 

2.475 

.423 

.906 

.466 

69° 

.934 

.358 

2.605 

.438 

.899 

.488 

70° 

.940 

.342 

2.747 

.454 

.891 

.510 

71° 

.946 

.326 

2.904 

.469 

.883 

.532 

72° 

.951 

.309 

3.078 

.485 

.875 

.554 

73° 

.956 

.292 

3.271 

.500 

.866 

.577 

74° 

.961 

.276 

3.487 

.515 

.857 

.601 

75° 

.966 

.259 

3.732 

.530 

.848 

.625 

76° 

.970 

.242 

4.011 

.545 

.839 

.649 

77° 

.974 

.225 

4.331 

.559 

.829 

.675 

78° 

.978 

.208 

4.705 

.574 

.819 

.700 

79° 

.982 

.191 

5.145 

.588 

.809 

.727 

80° 

.985 

.174 

5.671 

.602 

.799 

.754 

81° 

.988 

.156 

6.314 

.616 

.788 

.781 

82° 

.990 

.139 

7.115 

.629 

.777 

.810 

83° 

.992 

.122 

8.144 

.643 

.766 

.839 

84° 

.994 

.105 

9.514 

.656 

.755 

.869 

85° 

.996 

.087 

11.430 

.669 

.743 

.900 

86° 

.997 

.070 

14.300 

.682 

.731 

.933 

87° 

.998 

.052 

19.081 

.695 

.719 

.966 

88° 

.999 

.035 

28.636 

.707 

.707 

1.000 

89° 

.999 

.018 

57.290 















SUPPLEMENTARY EXERCISES 
TIMED TESTS 
SUMMARY 


SUPPLEMENTARY EXERCISES 


EXERCISE 215 

1. By the formula A = 1C, find l when A= 422.1, and 
C = 23.45. 

2. By the formula S = 2 irrh, find r when S = 1188, t = 2 T 2 -, 
and h = 21. 

3. By the formula A = \ a(b -f c), find a when A = 135, 
b = 15, and c = 21. 

4. By the formula 8 = \ gt 2 , find g if S = 257.6 and t = 4. 
(Remember that t 2 = t • t.) 

5. By the formula A — J ah, find a when A = 403 and 
b = 31. 

6. By the formula V = Iwh, find h when V = 3960, l = 
and w = 5. 

7. By the formula V = ^ 7rr 2 ^, find A when V = 264, 
7r = and r = 3. 

EXERCISE 216 

1. Find the area of the circle whose radius is 8 in. 

2. What is the area of the trapezoid whose altitude is 12 ft., 
whose upper base is 5 yd., and whose lower base is 7 yd. ? 

3. What is the simple interest on $1650 at 7% for 2 yr. 4 mo. ? 

4. What is the interest on $2500 at 6% for 3 yr. 9 mo. ? 

6. F — W/r is a formula used in physics. 

a. Find F when W = 250 and r = 5. 

b. Find W when F = 700 and r = 6. 

6. K = ^ Mv 2 is a formula used in physics. 

Find K when M = 200 and v = 25. 

7. H = .24 c 2 /ft is a formula used in physics. 

Find 1/ when J = 120, R = 12, and c = 10. 

426 


SUPPLEMENTARY EXERCISES 


427 


8. C = 


E 


is a formula used in the study of electricity. 

{K -f r) 

Find C when E = 1.08, R = 220, and r = 10. 

/ T \ mn 

9. S = PI 1 -\— ) is a formula used in investment mathe- 
V m) 

matics. 

Find the value of S when P = $2000, r = 6%, m = 2, and 
n — 1. 

L - l 


10. K = 


is a formula used in the study of heat. 


l(T-t) 

Find the value of K when L = 232, l = 230, T = 100, and t 

= 10 . 

11 . What is the volume of the sphere whose radius is 10 in. ? 

12. What is the area of the circle whose diameter is 15 in. ? 


EXERCISE 217 

Example 1. Add ax z and 2 bx 3 . 

Solution. — 1. These are like terms with respect to x 3 since they 
have the common factor x 3 . The coefficients of x 3 are a and 2 b. 

2. Therefore, by the rule on page 42, 

ax 3 + 2 bx 3 = (a + 2 6)x 3 . 

Example 2. Add h(x + y) and — 3(x + y). 

Solution. — These are like terms with respect to (x + y)> 

The coefficients of (x + y) are 5 and — 3. 

2. .*. 5(x + y) + [ — 3(x + y)] = (5 — 3)(x + y) = 2(x + y). 

Find the sum of: 


1. 

ax, bx, and cx 

3. 

2 ax 2 , ■ 

- 3 bx 2 , cx 2 , — 

4 (fa 2 

2. 

mx, nx, and — hx 

4. 

5 mx, - 

-2 nx, —6 px, + 8rx 

6. 

2(z + y), - 5(z + y ), 

and 8(x + y) 



6. 

— 3 (m + n) 2 , + 4 (m + n) 2 , 

— 5(m + n) 2 


7. 

Add 3 {x — y) — 4(x 

- z ) + % - 

z), - 2(x - 

y) + 

- 

z ) — 3(y — z), and (x 

- y) - 

- 2(x - 

z) + 4(y - z) 





428 


ALGEBRA 


8. Add 6 (a: + y ) 2 - 5 (a; + y) - 4, - 4{x + y) 2 -f- 2{x -f y) 
+ 3, and + 3(z + y) 2 - (x -f y) - 2. 

9. Add 8(r + s) + 5(£ + s) — 2(r -f <), — 7(r + s) -f- 

5(< + $) + 3(r + t), and 2(r + s) - 9 (t + s) - 5(r + t). 

10. Add ax + by + cz and dx — ey + fz. 

11. Add 2(x - y) + 3(a; + y) and 4(a; — y) — 5(x + y). 

12. Add 7 rs + 6 mn + 8 t and - 2 rs + 3 mn — 10 t. 

13. Add 13(a — a;) — 4(6 — x) + 3(c — x) and 2(a — x) + 
5(6 — x) — 7 (c — x). 

14. Add 2 ax 3 by + 4 cz and dx — 5 ey + 3 fz. 

15. Add ax 2 + bxy + cy 2 and Ax 2 — Bxy -f Cy 2 . 

16. Add 7 z 3 " - 5 x 2n + 3 x n - - 2 z 3 ” + 5 - 3 x n + 

6 x 2n , and — 5 x 3 ” — 4 z n + 8 z 2 ” — 7. 

EXERCISE 218 

1. From ab — be ac take ab be — ac. 

2 . From 4 z 3 — 10 x 2 — 18 + 11 x subtract 3 z 3 - 17 x + 

7x 2 - 20. 

3. Subtract 3z 3 -8+5z-6z 2 from 5 a: 2 - 11 + 8 x 2 

— 7 x. 

4. Subtract 2 x 2 — 3 xy + y 2 from 3 x 2 — 2 xy + y 2 . 

6. Subtract 12 a 3 — 4 a — 9 from 4 a 3 -f- 8 a 2 — 7. 

6. From x 5 - 4 x 4 y +3 z 3 ?/ 2 - 2 x 2 y* + xy 4 - y 5 subtract 

- 2 x 4 y -2 x b + x z y 2 - 3 x 2 y 3 - xy 4 + y b . 

7. From f a - T 7 ^ 6 - ^ c take | a + -^ 6 — £ c. 

8. Subtract - |v+iw-|a: from | v _i. w _j_5 a . 

9. How much more than 2 z -3?/ + 4zis5z - 6 i/ + 2 z ? 

10. How much is 2m— 3 n 4- 11 & greater than m — 2 n 
+ 4 jfc? 

11. From the sum of 3 a 2 - 5 a6 - 6 6 2 and 2 a 2 + 7 ab 
+ 4 6 2 subtract a 2 — 2 a6 -f- 6 2 . 


SUPPLEMENTARY EXERCISES 


429 


12. From 9 x 2 — 3 x + 5 take the sum of 2 x 2 4- 3 x — 4 
and 3 x 2 — 7 x + 6. 

13. From the sum of x — y — 3 z and 2 x — 8 z take the 
sum of 5 y — Ax + 7 z and 3 x —2 y —5 z. 

14. How much more than 3 a — 5 b is 10 ? 

15. Subtract f r — f 5 + 4 t from 1-J r + I s ~ tu 


EXERCISE 219 

1. From 3 ax take 2 bx. 

Solution. — 3 ax and 2 bx are like with respect to x. 
After the sign of 2 bx is changed, the coefficients of x are 
3 a and —2 b. Using the rule on page 42, the result is 
(3 a — 2 b)x. 

2. Subtract ay from by. 

3 . Subtract 3 mx from 4 nx. 


3 ax 
2 bx 


(3 a — 2 b)x 


4. Subtract ax + by from cx + dy. 

5. Subtract ax — by from cx + dy. 

6. Subtract ax 2 + bxy + cy 2 from Ax 2 + Bxy + Cy 2 . 

7. From mx 2 — nx + p subtract rx 2 — nx + p. 

8. From ax -f by — c subtract Ax + By — C. 

9. Subtract 2 ax + 3 by — 4 cz from 5 Ax + 4 By + 6 Cz. 

10. Subtract 5(x + y) 2 + 6(a + y) + 9 from 13 (x + y) 2 — 
3(x + y) - 5. 

11. Subtract 2(r + s) — 3 (t + r) — 6 {t + s) from 6(r -f s) 
— A{t + r) - 8 (t + s). 

12. Subtract 2(a + 6) 2 — 3(a + &)(« — &) + (a — &) 2 f rom 
3 (a + b) 2 + 4 (a + 6) (a — 6) — (a - fe) 2 . 

13. From 3 x 2n — 2 x n i/ m + i/ 2m subtract 2 x 2n + x n 2 / m - y 2m . 

14. From x 3 ” — x 2 ” + x n subtract x 3 ” — 2 x 2 " — x n . 

16. From 2 x* - x^y* + subtract x* + x^y* + V J • 




430 


ALGEBRA 


EXERCISE 220 

1. If A = P + PRT : 

a. Find R when A = $5000, P = $4000, and T = 3 yr. 

b. Find P when A = $3900, R = 6%, and T 7 = 5 yr. 

2. If Z = a + (n — l)d : 

а. Find d when Z = 40, a = — 5, and n = 10. 

б. Find a when Z = 197, n = 43, and d = 4. 



a. Find w if 8 = 750, a = 16, and l = 34. 

5. Find ti if S = 1000, a = 28, and l = 52. 

4. If y = a# + 5 : 

а. Find y when a = 7, x = 8, and b = — 12. 

6. Find 6 when y = 15, x = 2, and a = 4. 

б. If 8 = at — ^ gt 2 : . 

а. Find a when 8 = 3500, t = 8, and g = 32. 

б. Find a when 8 = 3500, t = 10, and gr = 32. 

6. If T = 2Trr 2 + 2Trrh (t r = 3|) : 

a. Find T when r = 14 and ^ = 20. 

b. Find h when T = 836 and r = 7. 

c. Find h when T = 704 and r = 7. 

7. If T = 7rr 2 -f- 7r rl : 

a. Find T 7 when r = 7 and Z = 10. 

Z>. Find Z when T 7 = 286, tt = - 2 /, and r = 7. 

EXERCISE 221 

Simplify: 

1. 6 c — J8 c — (10 c — 5) — 9| 

2. (8 Z - m) - [9 t - (5 t - 3 m) - 4 Z] 

3. (4 z — 2 y) — (6 * — [5 * + 7 y — 2] — 3 y) 

4. 12 c - J6 c — [9 c + 4] + [5 — 6 c]j 

5. 2 n - [3 n - j4 n + 4j - \ - 5 n - 9}] 


SUPPLEMENTARY EXERCISES 431 

6. a - (2 a - [3 a + 2] - [4 a - 5]) 

7. c — [2 c — (6 a — b) — (c — 5 a — 2 6)] 

8. 3 n — J4 n — (2 n + [5 n + 6]) j 

9. 5 r — \Q r — (7 r — [3 r — 2])| 

10. (8 t - 4) - [(5 t + 1) + (3 t - \2 t ~ 10] 

11. (11 ^ - 6) - [(3 z - 2) - (4 z + {6 z - 80] 

12. (2 r - 5) - l — (3 r + 2) - (7 r — [8 r — 1]) j 

13. m — [n + (3 m — 4 n)] — [2 m — 3 n — (m + 3 n)] 

14. (2 x - 3 y) - (4 y - [5 x - y] + [6 x - 4 y]) 

15. 3 ab — j (5 ab — 8) — (4 ab — 16) + (2 ab — 7) j 

EXERCISE 222 

Find: 

1. (z 8 -f- x 2 y + xy 2 + 3 /®)(z - y) 

2. (m 2 + % 2 ) (m 2 — n 2 ) (m 4 + w 4 ) 

3. (x - 3 ?/) 2 - (z + 3 t/) 2 5. (3z - 5?/) 2 - 5y(x + 5^) 

4. (a — z) 3 — (a + z) 3 6. (z + a + 6)(z — a — 6) 

7. (z 2 -4z+4)(z 2 +4z+ 4) 

8. 2(a + 2 z)(a — 2 z) - (a + 2z) 2 

9. (z 2 + z + 1) • z n 11. (a n — b n )(a n + b n ) 

10. (z n + 3)(z n + 3) 12. (z° + i/ 6 ) (z° + 2/0 

13. (z 3 " + 3 x 2n + 3 z n + 1) * x n 

14. (z 2a + z°.+ l)(z° — 1) 

15. (z 2 ° + z° + l)(z 2a — z° + 1) 

16. (z a - ?/ 6 ) (®° 4- y b ){x 2a 4- 2/ 26 ) 

EXERCISE 223 

Divide: 

1. 2 a 4 — 18 a 2 — 7 a 3 - a 4- 3 by 2 a 4- 1 

2. 10 a 3 4- 33 a 2 - 52 a 4- 9 by 2 a 2 4- 7 a - 9 

3. z 3 — 5 x 2 y 4- 9 xy 2 — 9 y 3 by z — 3 y 


432 


ALGEBRA 


4. m 4 + m 2 n 2 + w 4 by m 2 -f mn + n 2 
6 . m 3 + 2m 2 -9-6mbym 2 -m-3 

6. 8 — 64 y 3 by 2 a; — 4 y 

7. a 2 — 2 a& + 6 2 — c 2 by a — b + c 

8. 8 n 4 - 50 ?i 2 + 32 by 4 ?i 2 + 6 n - 8 

9. 6 z 4 + 13 z 3 - 70 z 2 + 71 z - 20 by 3 z 2 - 7 * + 4 

10 . a 5 —6a 3 + 6a 2 — 7a + 6bya 3 — 2a 2 + a—2 

11 . r 2 — s 2 — 2 ^ — t 2 by r — s — t 

12 . 8 a 2 + 60 be — 18 6 2 - 50 c 2 by 4 a + 6 b - 10 c 

13. a 5 — a 4 b — a6 4 + b b by a 2 — 2 a& + 6 2 

14. 8 m 5 — 14 m 2 — 18 m + 21 by 4 m 3 + 6 m — 7 
16. a 3 — 3 abc + 6 3 + c 3 by a + 6 + c 


EXERCISE 224 

Simplify by performing the indicated operations: 


E 

- + r 


4. x 2 + xy -b y 2 + 




3. a — 4 + 


7. 3 c + 1 - 


2 + 11 a 
3 a 

1 - 5c 


6. 4 m — 3 — 


6. m — n — 


16 m 2 


2 y 3 
x - y 
10 


4m + 3 


m 2 + mn + n 2 


4c - 3 


8. 

2m 

m 2 + m — 6 

9. 

+ 

00 

11 - 


l io 

10. 

l-2+ 2 +~2 1 


1 y 2 x 2 \ 

11. 

(«+ 6 f + ; ! i 


\ y 2 - l / 


3 m 


m 2 + 3 m — 10 
F 














SUPPLEMENTARY EXERCISES 


433 


12. (m + 1 — -7- (m — 4 — 

\ m/ \ mJ 

13. (a - -^t) + (a + — ) 

\ a — b) \ a + 2b) 

14 ' -ZT + O + ~^~h) 

a + b \ a — b/ 

15 . ( a - a ~ x ) 1 + ( a2 - aa: > \ 
\ 1 + ax' \ 1 + ax) 


16. 

\3 y 2x) \ 2 j x) 

17. ( 2x - 3+^4') -H fc* - 

V 2x- 3/ \ 7 4a:-6/ 


18. 


20 . 


21 . 


a — t _ a-f r 
a + r a — t 
2 _ 

3a: 4- 2 
5 x + 4 


f - 


(a - t)(a + r) 

a: — l) (a: 2 - 
x — 16' 


3a: +2 


■* 

■O+^^O+SeSH 


- -) 

x - 2/ 


r a: 2 + 6 a: + 8 > 
a: 2 + a: , 

m 2 ■+■ 6 mn +9 n 2 . /m 2 — 9 n 2 . m? — mn — 6 n 2 


m 2 + 4 mw +4 n 2 


/vr — 
‘ \m 2 - 


4 n 2 m 2 + mn — 6 n 2 ) 


1. 


EXERCISE 225 

7 a: 2 + 10 a: - 24 5 


(* + l) 5 


x + 1 


= 7 


n 6a: + 7_ , 0 _ 7a: 13 

2 -~6 X + ^ 2(2 x + 1) 


3. 


2 x — 1 _ 4 a: + 5 
3a: - 4 6 a: - 1 


4. 


6. 


5 a 4" 1 _ q 
2 q - 3 “ 2q - 3 


-3 


5 — x 


x — 3 3 — a: 


= 0 


6 . 


7. 


3a: - 4 
3 x 


_4_ = 1 

2a:—6 5a: — 15 2 

9 _1_ = 2 

3 q 5 u — 3 q — 








































434 


ALGEBRA 


y 


x — 2 3 + 4 2 3 — 1 

3 a _ 3 = 7 

2 a — 5 2 3a + l 


10. £ + — 

3 3 y - 


_ 9y -2 
4 9 


11 . 


12 


4 — 3 


3—3 1 — x 


+ 1=0 


12 . 


13. 


14. 


15. 


16. 


17. 


18. 


19. 


3 cl + 1 ci + 2 


= a 


3 2 a + 5 

5t 2 + 4t-7 St-2 


5 2+8 

3 a. ^ s — 4 
7 + 3x + l 

4 z — 36 


3 

2x - 5 


14 


11 


11 


x 2 — 9 a: — 3 
6 a — 6 


3 + 3 
26 - a 2 


1 


a — 4 a + 3 

y ~ 7 _ 

2/ 2 + 3 i/ — 28 ' y - 4 
17 9^ + 17 


12 


y + 7 
13 


2 3 — 16 3 2 — 2 3 — 48 2 3 + 12 

3 + 2 3 + 5 3+3 3 + 6 


3+3 3 + 6 3+4 3 + 7 

Combine the fractions in the first member; also those in 


Hint. 

the second member. Then clear of fractions. 


20 . 


y - 4 y - 1 
y - 5 y - 2 


y - 5 

2 / — 6 


y 


— 9 


2 / -3 


Exercise 226 — Literal Equations 

Solve the following for 3: 


i 3 + 5a 3 + a_ Q 
2 4 

x-7o_ y+ 8o_ 2c 


2 . 


10 


„ 3 + llr 10 r — 3 

6 3 

A X + 11 t \ j _ 10 t — 3 

4. - r t - —- 

















































SUPPLEMENTARY EXERCISES 


435 


6 x — 2 a _ 4 a 0 


6 . 


5 x x 
5 z -f 4 m 11 a: — 2 m 


2x 


6 a: 


= o 


10 . 


8 . 

— + a(h - 
9P 

Nn x 


Mm p — x 

13. 


3 x — 5 A 

9 x — 

7 A _ 2 

4 

12 

3 a: 

5 x — 4 & 

10 a: + 9 h _ 51 

5 

10 

6 a: 

cQ) = aA 

11 . 

4 a — x 



a — x 


12 . 

5 a: + 6 

a: 

2 x — Sb 

6 a: +'5 c 

2 c 

3 a: 

2 x 2 - 2 cx 

z 2 - , 

T.I 

1 

H. 

1 


12 a 


3 a — x 
x 


14. 


15. 


16. 


17. 


2 x _ 4 x + 5 t St 


3 a: — 4 t 6 x — t 3 a: — 4 t 

9 2 = _ 1 

S x — 5 r x — 2 r x — Sr 

3 4 _1__ 

a: — 2 s 2 x — s a: + 45 

ax . bx .7 

—rr ■I—i— = a + b 

x + b x + a 


18 a + b = J) + a 


x x — b x + a 

19 x \ 2 x 2 _ 9 x —2m 

3 3 x — 4 m 9 

5 x + t 4 a: + 7 t 3 x — 2 t 


20 . 


5 t 5 a: -+ 8 2 St 




































436 


ALGEBRA 


Exercise 227—Quadratic Equations 


1. 

2 . 

3. 


—— 4 - — = 0 
x — 4 5 £ 


15 c + 22 

2 +c 

3 


16 


2 - c 


-3 5-2 


4 2y + 11 _ y 4-17 
6 y - 21 5 y - 15 

6 . 6 = I+_6_ 

m 5 m + 1 



7. 3 - 4 *~ 4 

a; — 1 3 2 x 


8. 

2 - a ~ 

- 3 _ 3 a 

a - 

- 1 2a + 5 

9. 

2 - St 

2 _ 7 

4 

2-2 4 

10. 

6 

, § _ 3 — 2 v 

4 v — 3 

2v 

11. 

a 

2 a 4~ 5 

2a — 5 

3 a -J- 5 

12. 

1 

1 _ 4 

y- 2 

2/4-2 3 y 

13. 

7 

1 8 x - 12 

X +2 “ 

1 

II 

ICO 

14. 

2 w + 3 
5 

_ O w - 6 

w — 4 


1. 


2 . 


3. 


Exercise 228—Simultaneous Linear Equations 


3 m .n _ 17 
T" + 4 “ 8* 
m . n _ 4 
3 ^12 ” 3 


3j/ 

14 


= - 1 


+ = 8 
2 6 

x + 5y 2 ~ 3 .= 5 


x — 


4 y 


= -2 


4. 


6 . 


m _ 12 n — 11 
~2 10 

m . 22 n — 5 
■3 + 9 

g 2-4 k_ 


= - 2 

= - 1 
- 1 


1 i ,7 _ 5 1 ^_ _9 

12 + 2 


2x — 1 _ 5y + 1 
2 3 

Sy — 2 7x — 5 
2 5 


= 0 


r — 5 — 6 _ 7 _ r + 3 
2 6 3 

r + 3—9 s — r — 6 


9 


3 


to I CO 


















































SUPPLEMENTARY EXERCISES 


437 


11 . 


14. 


15. 


2 a + 3 6 2 a — 3 


5 

8 a + 5 6 


2 

10 a - b 


= 0 
= - 4 


r - t 


3r + 5f r Q 

~6 - 3" 8 

r -±l = ii _ 


10 9 . 

7"5 = 4 


15 

d 


x, 2y _7 a 

2 3 6 

* ~^y. = b 

3 6 

* - V = 9 
b a 

x _j_ y _ 3 b 2 — a 2 
a b 2 ab 

' x_±_2 _a , y = j 
a6 6 



10 . 


' 2x - 3 y _y_ 

6 3 

X - 3 _ x-4y = _ 17 



* + JL = i 

a& 6 


, x — y = a + 6 


Exercise 229 — Information Problems 

/'Wsf Degree Equations 

1 . The greatest depth of Lake Superior is about J that of 
Lake Michigan; the greatest depth of Lake Huron exceeds 




































438 


ALGEBRA 


one sixth that of Lake Michigan by 605 feet. The depth of 
Lake Superior exceeds that of Lake Huron by 265 feet. Find 
the depth of each. 

2. Probably the largest room in the world under one roof 
is the passenger concourse of the Union Station in Washington, 
D.C. Its perimeter is 1780 feet. One fifth of its length ex¬ 
ceeds its width by 22 feet. Find its dimensions. 

3. Ten times the population of the United States in 1820, 
in millions, exceeded the population in 1910 by 3.8 millions; 
the population in 1910 exceeded 7 times the population in 1820 
by 25 millions. Find the population in both years. 

4. Plants feed upon certain plant foods present in the soil, 
such as potash, nitrogen, and phosphoric acid. A fair crop of 
potatoes will remove from an acre of ground about 99 pounds 
of these three foods. The amount of potash removed is 5 
times, and the amount of nitrogen is 2\ times that of phos¬ 
phoric acid. Find the number of pounds of each used. 

6. The distance from San Francisco to London via New 
York is 6990 miles. The part of the journey by rail is 50 miles 
less than Ff of the part by water. Find the part of the journey 
on land and the part on water. 

6 . The elevation of Mt. Whitney, in California, the highest 
point recorded in the United States, is 14,501 feet, measured 
from sea level. The lowest point of dry land in the United 
States is in Death Valley, California. If 52 times the elevation 
of Death Valley be diminished by 45 and the result be increased 
by the elevation of Mt. Whitney, the sum is zero. Find and 
interpret the elevation of Death Valley. 

7. If the number of states admitted to the Union since its 
formation by the 13 original states is diminished by 9, the 
quotient obtained by dividing that number by 2 equals the 
original number of states. Find the present number of states 
(1912). 



SUPPLEMENTARY EXERCISES 


439 


8. I buy some bulbs from a seed store for $3, paying 75^ 
per dozen for one variety, and 50^f per dozen for another variety. 
The number of the first variety purchased exceeds the number 
of the second variety by 18. Find the number of each variety 
purchased. 

9. On the tower of the City Hall of Philadelphia is a statue 
of William Penn. If the total height of the tower and statue 
be divided by the height of the statue, the quotient is 14 and 
the remainder is 29; the height of the statue exceeds T V of the 
height of the tower by 7 feet. Find the height of the tower and 
of the statue. 

10. The City Hall of Philadelphia is said to cover a greater 
area than any other building in the United States. One fifth 
of its width exceeds one sixth of its length by 13 feet; and one 
ninth of its length exceeds one tenth of its width by 7 feet. 
Find its dimensions. 

Exercise 230 — Interest Problems 

First Degree Equations — One Unknown 

1. One sum of money is invested at 5%; a second sum is 
invested at 6%. If 3 times the first sum exceeds the second 
sum by $100, and if the total income is $155, find the sums 
invested. 

2. A man has $5000 invested at 4%. How much money 
must he invest at 6% to make the total income equivalent to 
5% on the total amount invested? 

3. A man has $3000 invested at 3.5%, and $4500 at 4%. 
How much must he invest at 6% to make the total income 
equivalent to 5% on the total sum invested? 

4 . A man has $3000 invested part at 5% and part at 6%. 
His total income per year is $157. How much has he invested 
at each rate ? 



440 


ALGEBRA 


5. The income at 5% on one sum of money exceeds by 
$35.50 the income at 4% on a sum which is $350 less than the 
first. Find the two sums invested. 

6. The income on one sum of money at 4j% and the income 
on a sum $600 greater at 3\% together amount to $421 per year. 
Find the total amount invested. 

Hint. —% = f % = 7 ^. 

7. A man invests a sum of money in 4§% stock and a sum 
$180 greater than the first in 3J% stock. If the incomes from 
the two investments are equal, find the sums invested. 

8. A man has $8000 invested at 5% and $5000 at 7%. 
How much must he invest at 8 % to make his income equal to 
6 % on his total investment ? 

First Degree Equations — Two Unknowns 

9. A man has $2500 invested from which he receives a 
total income of $135. Part of the money is invested at 6% 
and part at 4j%. How much is invested in each way? 

10. A man has all together $5000 of savings. He has part 
invested in a 5% bond, and the balance invested in a mortgage 
drawing 6%. If his total income is $280, how much has he 
invested in each way ? 

11. A man has $1200 invested at one rate of interest and 
$500 at a rate which is 1 per cent greater than the former rate. 
The income from the first investment exceeds the income from 
the second investment by $23. Find the rate at which each 
sum is invested. 

12. The simple interest on $800 at 5% for a certain number 
of years exceeds the simple interest on $300 at 6% for a second 
period of years by $60. If the second period of years exceeds 
the first by 4 years, find the number of years each sum is 
invested. 


SUPPLEMENTARY EXERCISES 


441 


Exercise 231 — Age Problems 

First Degree Equations — One Unknown or Two Unknowns 

1. A father is now 9 times as old as his son. In 9 years he 
will be only 3 times as old as his son. What are their present ages ? 

2. The difference between the present ages of a father and 
son is 25 years. In 10 years the father will be twice as old as 
his son. What are their present ages ? 

3. A is 5 times as old as B. In 9 years he will be only 3 
times as old as B. What are their ages ? 

4. B is twice as old as A. 35 years ago he was 7 times as 
old as A. What are their present ages ? 

6. A is 68 years of age, and B is 11. In how many years 
will A be 4 times as old as B ? 

6. The age of a father is 5 times that of his son; his age 5 
years from now will exceed 3 times his son’s age by 4 years. 
Find their present ages. 

7. A’s age is three eighths of B’s, and 8 years ago it was two 
sevenths of B’s age. Find their ages at present. 

8. Washington was admitted to the Union 18 years before 
Oklahoma, and may therefore be said to be 18 years older than 
Oklahoma as a state. One fourth of Washington’s age in 1911 
exceeded Oklahoma’s age by 1| years. Find the year when 
each was admitted. 

9. In 1912, the “ age ” of Maine was 4 T 2 T times that of 
Wyoming; in 1920, it will be 3^ times it. Find when each 
state was admitted to the Union. 

10 . A’s age 11 years from now divided by his age 11 years 
ago is the fraction Find his present age. 

11. A’s age exceeds twice B’s age by 7 years. The sum of 
three fifths of B’s age 2 years ago and of four sevenths of A’s 
age 4 years from now is 26 years. Find their present ages. 


442 


ALGEBRA 


12 . A’s age is three fifths of B’s age; but in 16 years A’s age 
will be five sevenths of B’s age. Find their ages at present. 

13. Three years ago A’s age was ^ of B’s age; but in nine 
years his age will be of B’s age. Find their present ages. 

14. If the age of a university is reckoned from the date of its 
founding, then, in 1912, the age of Yale exceeded the age of 
Princeton by 46 years. Four years later, one half of the age of 
Princeton will exceed one third of the age of Yale by 13 years. 
Find when each was founded. 

15. In 1912 the sum of the ages of the oldest English uni¬ 
versity, Oxford, and the oldest American university, Harvard, 
was 1316 years; the age of Oxford exceeded three and three 
fourths times the age of Harvard by 5 years. Find when each 
was founded. 

Exercise 232 — Geometry Problems 

1. In a triangle commonly used by draughtsmen, the second 
angle is two thirds of the first, and the third angle is one half of 
the second. Find the angles of the triangle. (§ 13) 

2. In another triangle used by draughtsmen, there are two 
equal angles, each of which is one half of the third angle. Find 
the angles of this triangle. 

3. Probably the largest room in the world under one roof is 
the passenger concourse of the Union Station in Washington, 
D.C. Its perimeter is 1780 feet. One fifth of its length ex¬ 
ceeds its width by 22 feet. Find its dimensions. 

4. The length of the foundation of the Capitol in Washing¬ 
ton exceeds twice the width by 51£ feet. The perimeter of the 
foundation is 2202|- feet. Find the dimensions of the founda¬ 
tions of the Capitol. 

6. Twice the width of the Pennsylvania Station in New 
York exceeds its length by 80 feet. Four times the length 
exceeds the perimeter by 700 feet. Find the dimensions. 


SUPPLEMENTARY EXERCISES 


443 


6. Find the three sides of a triangle if the perimeter is 45 
inches, if the second side is twice the third side, and if the first 
side exceeds the third by 5 inches. 

7 . The base of a certain rectangle exceeds its altitude by 8 
inches. If the base and altitude are both decreased by 2 inches, 
the old area exceeds the new by 36 square inches. Find the 
dimensions of the rectangle. 

8. The base of a rectangle is 9 feet more and the altitude is 
8 feet less than the side of a square. The area of the rectangle 
exceeds the area of the square by 15 square feet. Find the 
dimensions of the rectangle. 

9. A man planned a house whose length exceeded its width 
by 10 feet. He found that it would be too expensive to build 
the house as planned, so he decided to decrease both dimensions 
5 feet. This decreased the area of the house 425 square feet. 
What were the original and the new dimensions ? 

10. A man planned to set out an apple orchard with two 
more trees to each row than he had rows, but found that that 
plan left one tree over. He found that if he decreased the 
number of rows by 3, and increased the number of trees per 
row by 5, he used all of his trees. How many trees had he ? 

Quadratic Equations — One Unknown 

11 . The length of a certain rectangular farm is three times 
its width. If its length should be increased by 20 rods, and 
its width by 8 rods, its area would be trebled. Of how many 
square rods does the farm consist? 

12. The standard size city lot in parts of Chicago is five 
times as long as it is wide. The lots in parts of Indianapolis 
are 10 feet wider and 5 feet longer than those in Chicago. Three 
times the area of a Chicago lot exceeds twice the area of an 
Indianapolis lot by 275 square feet. Find the dimensions of 
the lots in both cities. 


444 


ALGEBRA 


13. An architect who has made plans for a house with a base 
30 X 42 feet finds that he must reduce the size. By what 
equal amount must he reduce the two dimensions of the house 
in order to make the area of the new base 925 square feet ? 

14. Find the angle such that 3 times its complement increased 
by two thirds of its supplement equals 137°. 

15. The supplement of a certain angle divided by its comple¬ 
ment gives as quotient 2%. What is the angle ? 

16. The width of a room is three fifths of its length; if 12 
feet be added to the width and taken from the length, the room 
will be a square. Find its dimensions. 

17. Five lines radiate from a point making angles such that 
the second is one half of the first, the third is twice the first, 
the fourth is the sum of the second and third, and the fifth is 
three times the third. Find the angles. 

18. Find the three angles of a triangle if the second angle 
is one half of the remainder obtained by diminishing the first 
angle by 1°, and if the third angle is f of the remainder obtained 
by diminishing the first angle by 7°. 

First Degree Equations — Two Unknowns 

19. The City Hall of Philadelphia is said to cover a greater 
area than any other building in the United States. One fifth 
of its width exceeds one sixth of its length by 13 feet; and one 
ninth of its length exceeds one tenth of its width by 7 feet. 
Find its dimensions. 

20. The perimeter of a certain isosceles triangle (see p. 102) 
is 140 inches. Each side exceeds the base by 10 inches. Find 
the three sides of the triangle. 

21 . Twice the shorter side of a parallelogram exceeds the 
longer side by 5 inches; one third of the sum of the shorter 
side and 9 exceeds one fifth of the longer side by 3. Find the 
sides of the parallelogram. 


SUPPLEMENTARY EXERCISES 


445 


22 . If a rectangular lot were 6 feet longer and 5 feet wider 
than it is now, it would contain 839 square feet more; if it were 
4 feet longer, and 7 feet wider, it would contain 879 square feet 
more. Find its length and width. 

23. If a field is made 5 feet longer and 7 feet wider, its area 
would be increased by 830 square feet; but if its length is made 
8 feet less, and its width 4 feet less, its area is diminished by 700 
square feet. Find its length and width. 

24. There are two supplementary angles such that £ of the 
larger exceeds § of the smaller by 5°. Find the angles. 

25. One angle of a triangle is 35°. If the number of degrees 
in one of the remaining two angles is divided by the number in 
the "bther, the quotient is 9 and the remainder is 10. Find the 
three angles of the triangle. 

Quadratic Equations — Two Unknowns 

26. Find the dimensions of a rectangle whose area is 357 
square feet if its length exceeds its width by 4 feet. 

27. The main waiting room of the Union Railway Station 
in Washington, D.C., has an area of 28,600 square feet. The 
length exceeds the width by 90 feet. Find the dimensions. 

28. The area of a rectangular field is 216 square rods, and 
its perimeter is 60 rods. Find the length and width of the field. 

29. The altitude of a certain rectangle is 2 feet more than 
the side of a certain square; the perimeter of the rectangle is 
7 times the side of the square, and the area of the rectangle 
exceeds twice the area of the square by 32 feet. Find the side 
of the square and the base of the rectangle. 

30. Find the sides of a parallelogram if the perimeter is 24 
inches and the sum of the squares of the number of inches in 
the long and short sides is 80. 


446 


ALGEBRA 


31. One of two angles exceeds the other by 5°. If each is 
multiplied by its supplement, the product obtained from the 
larger of the given angles exceeds the other product by the 
square of the smaller of the given angles. Find the angles. 

32. Two angles are supplementary. The square of the num¬ 
ber of degrees in the larger angle exceeds by 4400 the product 
of the number of degrees in one angle by the number in the other 
angle. Find the number of degrees in each angle. 

33. A man has two square lots of unequal size, together 
containing 74 square rods. If the lots were side by side, it 
would require 38 rods of fence to surround them in a single 
inclosure of six sides. Find the length of the side of each.- 

Exercise 233 — Distance, Rate, and Time Problems 

1. Suppose that two men, who travel at the rate of 6 miles 
and 10 miles per hour respectively, start from the same place 
in opposite directions. In how many hours will they be 200 
miles apart? 

2 . Suppose that A, traveling 10 miles per hour, leaves a 
place 3 hours before B; suppose that B travels 15 miles per 
hour. In how many hours will B overtake A? 

3. Suppose A, traveling 15 miles per hour, starts 4 hours be¬ 
fore B. At what rate must B travel to overtake A in 10 hours ? 

4. Two hours after A left, B starts after him in an auto¬ 
mobile at the rate of 27 miles an hour and overtakes him in 2\ 
hours. At what rate was A traveling ? 

6. A and B travel toward each other from points separated 
by 250 miles, A at a rate which exceeds B’s rate by 8 miles an 
hour. If they meet in 5 hours, at what rate did each travel ? 

6. Some boys who are boating on a river know that they 
can go with the current 6 miles per hour and can return against 
the current at the rate of 3 miles per hour. How far may they 
go if they have only 3 hours for the trip ? 


SUPPLEMENTARY EXERCISES 447 

7. A man has 11 hours at his disposal. How far may he 
go in a buggy at the rate of 10 miles an hour if he plans to return 
at an average rate of 7 miles per hour? 

8. An automobile is traveling at the rate of 25 miles an hour. 
In how many hours will a second automobile overtake the first 
if the second starts 2 hours later than the first, and travels at 
the rate of 35 miles an hour? 

9. An express train whose rate is 36 miles an hour starts 
54 minutes after a slow train and overtakes it in 1 hour and 
48 minutes. What is the rate of the slow train ? 

10. An automobile party is traveling at the rate of 20 miles 
per hour. At what rate must a second automobile travel in 
order to overtake the first if it starts 2 hours after the first and 
wishes to overtake it in 3 hours ? 

11 . Chicago and Madison, Wisconsin, are about 140 miles 
apart. Suppose that a train starts from each city toward the 
other, one at the rate of 35 miles per hour and the other at the 
rate of 40 miles per hour. How soon will they meet ? 

12. The rate of an express train is three times that of a slow 
train. It covers 180 miles in 8 hours less time than the slow 
train. Find the rate of each train. 

13. A messenger starts out to deliver a package to a point 
24 miles distant, at the rate of 8 miles per hour. At what rate 
must a second messenger travel to arrive at the same time as 
the first messenger, if he starts 1 hour after him ? 

14. The rate of a passenger train exceeds twice the rate of 
a freight train by 5 miles per hour. It can go 350 miles while 
the freight train goes 150 miles. Find the rate of each train. 

16. Two men, A and B, 57 miles apart, travel towards each 
other, B starting 20 minutes after A. A travels at the rate of 
6 miles an hour and B at the rate of 5 miles an hour. How 
far will each have traveled when they meet? 


448 


ALGEBRA 


16. The rate of an express train is five thirds of that of a slow 
train. It travels 36 miles in 32 minutes less time than the slow 
train. Find the rate of each train. 

17. The rate of an express train is three halves that of a 
slow train. It covers 270 miles in 3 hours less time than the 
slow train. Find the rate of the trains. 

Linear Equations — Two Unknowns 

18. An express train travels 30 miles in 27 minutes less time 
than a slow train. If the rate of the express train were as 
great, and if the rate of the slow train were f as great, the 
express train would travel 30 miles in 54 minutes less time 
than the slow train. Find the rate of each train in miles an 
hour. 

19. The fastest train on the Pennsylvania R. R. makes the 
trip between Chicago and Fort Wayne, Indiana, a distance of 
148 miles, in 1 hour and 20 minutes less time than one of the 
ordinary trains. Its rate is f that of the ordinary train. Find 
the rate of each train. 

Quadratic Equations 

20 . A man traveled 105 miles. If he had gone 9 miles more 
an hour, he would have performed the journey in 1^ hours less 
time. Find his rate in miles an hour. 

21 . If a man travels 120 miles by one train and returns on a 
train whose rate is 10 miles an hour more, he will require 7 hours 
for the trip. What is the rate of the first train? 

22 . An automobile made a trip of 50 miles, 10 miles within 
city limits and 40 miles outside city limits. Outside city limits, 
the rate was increased 15 miles an hour. If the trip took 2f 
hours, at what rate did the car travel within and outside city 
limits ? 


SUPPLEMENTARY EXERCISES 


449 


Exercise 234 — River Problems 

Linear Equations — Two Unknowns 

1. Some boys who can row 4 miles an hour in still water 
made a trip on a river whose current is 2 miles an hour. If it 
took them 8 hours for the trip, how far did they go ? 

2. A party took a trip in a motor boat which runs at the 
rate of 15 miles an hour. They took three hours for the trip. 
What distance did they go, if the rate of the current is 3 miles 
an hour ? 

3. Some boys row on a river whose current is known to be 
2\ miles per hour. They find that it takes them as long to go 
upstream 2 miles as downstream 7 miles. What is their rate of 
rowing ? 

4. Some boys are rowing on a river whose current is 5 miles 
per hour in one stretch and 3 miles in another. They find when 
going downstream that they can go 4 miles where the current 
is rapid in the same time that they can go 3 miles where the 
current is slower. Find the rate at which they row in still 
water. 

5. A crew can row 10 miles downstream in 50 minutes, and 
12 miles upstream in an hour and a half. Find the rate in miles 
an hour of the current, and of the crew in still water. 

6. A motor boat which can run at the rate of 15 miles an 
hour in still water went downstream a certain distance in 4 
hours. It took 6 hours for the trip back. What was the dis¬ 
tance and the rate of the current? 

7. A crew is rowing on a stream the rate of whose current 
is known to be 2 miles an hour; they find that it takes them 
one and one third hours to go down, and four hours to come 
back a certain distance. Find the distance and the rate of the 
crew in still water. 


450 


ALGEBRA 


8 . A crew can row 8 miles downstream and back again in 

hours; if the rate of the stream is 4 miles an hour, find the 

rate of the crew in still water. 

9. Some boys are canoeing on a river in part of which the 
current is 5 miles an hour, and in another part 3 miles an hour. 
If, when going downstream, they go 4 miles where the current 
is rapid and 8 miles where it is less rapid in a total time of If 
hours, what is their rate of rowing in still water? 

Exercise 235 — Digit Problems 

1. The tens’ digit of a number exceeds its units’ digit by 
4. If the digits be reversed, the new number is 6 more than 
one half of the old number. Find the number. 

2. The sum of the two digits of a number is 9. If the 
digits be reversed, the quotient of the new number divided by 
the units’ digit of the given number is 13 and the remainder 
is 1. Find the number. 

3. The sum of the two digits of a number is 16; and if 18 
be subtracted from the number, the remainder equals the 
number obtained by reversing the digits. Find the number. 

4. If the digits of a number of two figures be reversed, the 
sum of the resulting number and twice the given number is 
204; and if the given number is divided by the sum of its 
digits, the quotient is 7 and the remainder is 6. Find the 
number. 

5. If a certain number be divided by the sum of its two 
digits, the quotient is 4 and the remainder is 3. If the digits 
be reversed, the sum of the resulting number and 23 is twice 
the given number. Find the given number. 

6. The sum of the two digits of a number is 15. If the 
digits be reversed, the new number exceeds two thirds of the 
original number by 35. What is the number ? 


TIMED TESTS 


Purposes. — To afford a ready means : 

a. Of comparing individual pupils with the class; 

b. Of comparing one class with another; 

c. Of stimulating pupils to work more rapidly, and to work 
more accurately when hurried. 

Note. — If the following sets of examples are not used as “timed 
tests,” they will be found useful as so many extra drill examples. 

How to be used. — 1. Urge pupils not to write any answers 
in the text. 

2 . Tell them to start promptly and to stop at once when the 
word is given. Explain the purpose of limiting the time. 

3. Tell them that both speed and accuracy are important. 
Tell them that it is not expected that many, or even any in some 
cases, will complete all the examples in the time allowed. 

4. Have the answers ready in advance so that you can give 
your whole attention to watching the pupils and the clock. 
Stop the class when the allotted time has expired. 

5. Have the papers marked at once by the pupils. If neces¬ 
sary, have the pupils exchange papers. Read the answers 
once, slowly and distinctly. 

6 . Have indicated on each paper: a. The number of examples 
tried. b. The number of examples right, c. The percentage 
which the latter is of the former. ( I.e ., if a pupil tried 15 ex¬ 
amples out of 20 examples, and got 10 right, his accuracy mark 
is of 100% (66.6%) and not -J# of 100% (50%). 

7. Derive the median speed of the class as follows: 

Direct all who tried only 1 to stand and remain standing; 

similarly those who tried only 2, only 3, etc., until one half the 
class is standing. The number of examples tried by the last 
pupil to rise is the median speed of the class for the allotted 
time. 


451 


452' 


ALGEBRA 


8. Derive the median accuracy percentage of the class as 
follows: 

Direct all who made 100% (see 6 above) to stand, and remain 
standing; similarly those who made 99%, 98%, etc., until one 
half the class is standing. The mark of the last pupil to rise 
is the median accuracy mark of the class. 

9. Record these medians on a sheet to be made out in duplicate 
(if you will) as follows : 


Record of Wells and Hart Timed Tests 


School. Teacher. 

Number of pupils. Teacher’s next summer address 


Test 

No. 

Date 

Given 

Median 

Speed 

Median 

Accur’cy 

Number of 
100% Pupils 

Date 

Given 

Median 

Speed 

Median 

Accur’cy 

Number of 
100% Pupils 

I 









II 









Etc. 

Provide space for twenty tests. 


Keep one copy yourself. At the close of the school year, send the 
other copy to Professor W. W. Hart, University of Wisconsin. Early 
in the summer he will send you a statement showing the averages of 
the medians for each test, as reported to him. 

10. Most important of all. — Assign special practice to all 
pupils who fall below the medians in speed or accuracy. Pro¬ 
vide opportunity for them to demonstrate that they have 
attained the medians of the class. 

Table I (Used in Tests I, II, and III) 


1. 

2. 

3. 

4. 

5. 

6. 

7. 

8. 

9. 

10. 

+ 3 

- 6 

- 2 

+ 6 

+ 8 

+ 3 

+ 2 

+ 8 

- 6 

- 3 

-9 

+ 4 

- 9 

- 3 

- 2 

- 7 

+ 7 

- 5 

- 5 

- 8 
























TIMED TESTS 453 


11. 

12. 

13. 

14. 

15. 

16. 

17. 

18. 

19. 

20. 

- 7 

-8 

- 6 

+ 4 

- 3 

- 7 

+ 9 

-8 

- 6 

- 8 

- 5 

+ 2 

- 3 

- 9 

+ 9 

- 2 

- 6 

+ 3 

- 4 

- 7 

21. 

22. 

23. 

24. 

25. 

26. 

27. 

28. 

29. 

30. 

- 5 

+ 9 

- 3 

- 7 

+ 9 

+ 7 

- 9 

- 7 

- 9 

+ 8 

- 8 

- 4 

- 7 

+ 6 

- 2 

- 9 

- 8 

±_8 

- 5 

- 4 

31. 

32. 

33. 

34. 

35. 

36. 

37. 

38. 

39. 

40. 

+ 6 

- 7 

+ 9 

- 9 

-8 

+ 5 

+ 8 

- 7 

- 9 

+ 9 

- 5 

- 9 

- 8 

- 6 

- 4 

- 7 

-8 

- 6 

- 9 

- 5 


Note. - For Tests I—III, have the pupils prepare in advance a sheet 
containing in one column the numbers 1 to 20, written about one half 
inch apart; and, in a second column, the numbers 21 to 40. Tell 
them they are to write beside each number the answer to an example 
having the same number. 


Test I — Addition of Signed Numbers 
Time: 3 Minutes 

Add the numbers in Examples 1 to 40. Write only the 
answers. 


Test II — Subtraction of Signed Numbers 
Time: 3 Minuted 

Subtract the lower number from the upper in Examples 1 to 
40. Write only the results. 

Test III — Multiplication of Signed Numbers 
Time: 3 Minutes 

Multiply the two numbers in Examples 1 to 40. Write only 
the products. 


454 


ALGEBRA 


Test IV — Evaluation of Expressions 
Time: 10 Minutes 


Find the numerical value of the following expressions when 


1. 3 a -f - 2 6 

2. 5 c - 3 d 

3. m 2 + mx 

4. z 3 — z 2 + 1 
6. 8 a - 3 6 

6. 9 c + 2 d 

7. 2m 2 + 3 mx 2 


6 = 5, c = 3, d = 4, m = 5, x = 2. 
8. 2 x 3 + 3 a; 2 - 5 


9. | a6 

10. \ a(6 + c) 

11 . 2 

12. 4 • V- • z 2 

13. f a6 2 

14. 2 cd + 3 be 


15. m 3 — m 2 x 

16. z 3 — 2 x 2 + x 

17. a 2 + 2 a6 

18. 5 c 2 — 3 cd 

19. 2 mx + 3 mx 2 

20. ^-2^ + a: 2 

21. ax 2 — bx + c 


Table II (Used in Tests V, VI) 

1. 7a—36+4 c, and 5a — 76+2c 

2. Qr — 7s-\-t, 8r—Ss, and 2 r + 5 t 

3. 5a; — 7?/— 9z, 3t/+4z, and — 2 x + 5 z 

4. — 3c + 4d, 9c — 10d + 3e, and — 4 c + 3 e 

5. 12 p — 7 r, — 8p + 5r — 3 5, and 7 p — 3 r — s 

6. x 2 — 3 a; + 6, — 2 a; 2 + 7, and 3 a; 2 + 5 a; -■ 8 

7. 8 y 2 — 2 y + 4, — 6 y — 7, and — 5 y 2 + 6 

8. 3 m 2 — 6 + 7 m, — 4 w + 10, and 5m - 7 

9. - 4 t 2 + 6 * - 9, 3 * 2 - 8 t + 2, and 2 t 2 - 3 < + 1 

10. 11 s 2 — 5 s — 3, — 4 s 2 + 6 s + 2, and — 3 s 2 — 8s+5 

11. 2 a: 2 — 3 xy + y 2 , — 3 x 2 — 2 2 / 2 + 5 xy, and — 2 ?/ 2 + 4 a; 2 

12. 5 m 2 — mn — n 2 , 4 ?i 2 — 2 m 2 + 3 mn, and —3 n 2 — m 2 

13. — pr + 4 r 2 , 3 p 2 — 2 r 2 , and 6 pr + 7 p 2 — 5 r 2 

14. — 10 ab + 6 a 2 — 5 6 2 , a 2 + 2 ab — 3 6 2 , and — 4 a6 + 5 a 2 

15. 9 c 2 — d 2 — 6 cd, 5 cd — 4 d 2 , and 3 d 2 — 7 c 2 + 2 cd 


TIMED TESTS 


455 


Test V Addition of Polynomials 
Time: 15 Minutes 

Copy and add the polynomials in parts 1 to 15 of Table II. 

Test VI Subtraction of Polynomials 
Time: 10 Minutes 

For this test, subtract the first polynomial from the second in 
each of the fifteen parts of Table II. Copy the polynomials 
before subtracting. 

Table III (Used in Tests VII, VIII) 

1. 2 w 2 - 11 w 5; 2 w - 1 3. 6 m 2 - 11 m + 3; 2m - 3 

2. 3 b 2 - 19 b + 6; b - 6 4. 21 ,4 2 + 2 A - 8 ; 3 A + 2 

5. 10 z 2 - 13 x - 30; 2 a: - 5 

6. 5c 2 -23cd + 12d 2 ; 5c - 3d 

7. 10c 2 + 9cd — 9 d?; 2 c -\- 3 d 

8. 4 x 2 - 15 y 2 + 4 xy ; 2 x + 5 y 

9. 4 a 2 — 21 b 2 — 8 ab; 2 a — 7 b 

10. 9 r 2 — 35 s 2 — 6 rs ; 5 s + 3 r 

11. x 3 — 5 x 2 + 9 x — 9; x — 3 

12. 4 a 3 + 8 a 2 - 39 a + 27; 2 a - 3 

13. 9 c 3 - 34 c - 9 c 2 - 16; 3 c + 2 

14. 2 m 3 - 7 m 2 - m + 15; m 2 - m - 3 

15. 2 z 3 - 2 x 2 y + xy 2 - 6 y 3 ; 2 x 2 + 2 xy - 3 y 2 

Test VII — Multiplication of Polynomials 
Time: 15 Minutes 

Copy and multiply the first polynomial by the second in 
parts 1 to 15 of Table III. Fifteen examples. 


456 


ALGEBRA 


Test VIII — Division of Polynomials — Time : 15 Minutes 

Copy and divide the first polynomial by the second in each 
of the fifteen parts of Table III. Fifteen examples. 


Test IX — Special Products — Time: 5 Minutes 


Do not copy the factors. Write only the products. 


1. (3* + 2y)(3a -2y) 

2 . (2 a - 3 6) (2 a - 3 b) 

3. (9 a - 2 y)(a + y) 

4. (; x 2 - 5 y)(a 2 + 5y) 

5. (6s - 5 0(2* +3<) 

6. (3 t + 4 n)( 3 t + 4 n) 

7. (9 ab — 4 c) (9 a& + 4 c) 

8. (7 r 2 + 8s)(3r 2 - 2 5 ) 

9. (x - 10 y)(x + 5y) 

10 . (7—6 xy) (5 + 6 ay) 


11 . (2 + 9 a) (3 - 2 a) 

12. (10 a 3 -3 y 2 ) (2 a 3 +4 y 2 ) 

13. (9 mn 2 + 4) (6 mn 2 — 3) 

14. (a 2 + 7 yz)(5 a 2 — 4 yz) 

15. (7 a 2 4- 3 6 2 ) (5 a 2 - 2 6 2 ) 

16. (3 c 2 - 5(0(4 c* -5d) 

17. (20a - 76)(8a + 3 6) 

18. (10 m -n)(6m + 5n) 

19. (8-3 a 2 ) (10 + a 2 ) 

20. (14 rs — 9) (2 rs + 1) 


Test X — Factoring — Time: 5 Minutes 


Do not copy the example. 

1. 2 m 3 — 10 m 2 

2. 16 a 2 - 25 y 2 

3. z 2 — 12 2 + 35 

4. 8 a 2 — 22 a + 15 

6. 2 a 2 + a — 15 

6. 9 r 2 — 6 r — 8 

7. 15 c 2 - 4 c - 3 

8. r 2 s — 4 rs —45 s 

9. 4 a 2 — 24 a + 35 

10. 3 z 2 + 22 z + 7 


Factor. 

11. m — 36 ma 2 b 2 

12. 4 a 2 — 28 ax + 49 a 2 

13. a 2 + 7 ay — 60 y 2 

14. 6 a 2 + 31 a + 35 

15. 12 a 2 + 13 a - 35 

16. 2 a 2 - 3 ay - 20 y 2 

17. s 2 — 6 st — 55 t 2 

18. 10 a 2 + 9 ab —9 b 2 

19. 81 a 2 — 72 ay + 16 y 2 

20. 4 a 2 - 8 xy — 21 y 2 


TIMED TESTS 


457 




Test XI — Simplification 
Time: 10 Minutes 


Copy the examples. Remove parentheses and combine terms. 

1. 3(2 x - 4) + 2{x - 5) 4. - 2 (t + 9) - 3 (t - 4) 

2. 6(2 — 3 x) — 3(4 x — 7) 5. — 4(c — 3) — 3(2 c — 5) 

3. — 2(a — 6) -J- 5(3 a — 2) 6. 2 x(x — 3 y) + y{x 2 y) 

7. 3 a(2 a - 4 b) - b(a -5 b) 

8. — 5 r(s — 2 r) + 2 s(r — 3 5 ) 

9. - 4 c(3 c + 2 d) - c(2 c - d) 

10 . 8 ra(ra — 3 n) — 5 n(2 m + n) 

11. (x — 4) (s + 6) — Or + 5) (2 — 2) 

12. (y + 3)(2 y — 1) + (3 ?/ — 2){y — 4) 

13. (2 x — l)(x + 3) - (3 * + 2)Or - 1) 

14. 2Or + 2)Or - 3) + S(x - 4)(* + 5) 

15. 3(2 * + l)Or - 2) - 2(3 * - 2){x - 3) 


Test XII — Simple Equations 
Time: 10 Minutes 


Copy the equation. Solve. 
1. 5 a + 5 = 61 —3a 
0 x x _ 5 

2 ‘ 2 3 “ 6 

3. 4(y - 5) + 7 = 15 

4. 9m — 7 = 3m — 37 

_ 2 s $ _lc 

6. T = i + 5 

6. 3 — Or — 3) = 5 — 2 £ 

7. 13 — 6 3 = 13 z — 6 

’ 3 m n m 

9. 3(x + 1) = 10(x — 6) 


Omit check. 

10 . 7t + 10 = 16 < — 17 

11 £-®=§-£_LS 

'2 3 4 3 

12. 4c = 6(3 - 4c) - 7(4 + c) 

13. 15 - 6 n = 5 n + 48 

14 i«-5J = 2_3^ 

'9 6 3 2 

15. (3 x + 4) — (2 x + 9) =15 

16. 13 + 4p = 11 p — 22 

17. 2*' _ » _ 52 _ 5 

3 8 6 4 

18. 3ra - 2(2re - 7) = 3(re-2) 







458 


ALGEBRA 


Test XIII — Fractional Equations 
Time: 15 Minutes 
Copy the equations. Solve them. 


x + 5 _ x -+■ 1 
2 4 

6 _ 1 = 5 
a a 3 

12 =3 


x — 2 

11 + rn _ 10 — m _ ^ 
* 6 3 

6 . *Jz2 _ 4 _ 2 

5 a; x 


10 


9 


m — 3 m — 5 

„ a + 5 „ a — 10 

7. — 1 -4 = 


3 4 

5* +4 in -2 


2 * 
a; 4- 5 
a; — 3 


6 t 


= 3 


= 5 


x — 3 


10 . 

11 . 

12 . 


c - 2 _ c - 4 = 2 
4 6 3 

r - 3 _ 1 _ 3r - 7 
2 r 3 2 r 


5 x + 1 _ x 
2a;—3 2 a; - 3 


13 .^i + 3 =^ 14 


-3 


3 9 

12 2 a;+8 


14. — - 

x 

4 — x 


5 x 


15. 


12 


1 — x 3 — x 


+ 3=0 


+ 1 


16 7 w + 1 _ w ~ 9 = 1 


17. 

18. 


4 4 

3 m — 5 _ 9m - 7 _ 2 
4 12 '3m 

2 _ 5 = 2 

a; + 2 x — 2 x 2 -4 


Test XIV — Simultaneous Equations 
Time: 15 Minutes 


Solve the odd ones by the addition-subtraction method of 
elimination; the even ones by the substitution method. 

4 a + 3 6 = —1 

5 a + b = 7 

3 m + 2 w = 17 4 ( r — 6 5 = 2 

4 m + 7i = 16 1 3 5 — 8 r = 29 


1 | 3x + y = 11 

5 x — y = 13 


'•! 



































TIMED TESTS 


459 


5. 

6 . 

7. 

8 . 

9. 


\r-6s= - 10 
l 2r - 7 s = - 15 
f 5 x -f 8y = 2 
1 10 x - 12 y = 32 
f5m+4n=22 
l 3 m + n =9 
f 2 a: + 5 y = 13 
\7 x — 4 ?/ = — 19 
f 7c - 2d = 31 
l 4 c - 3 d = 27 


8 x + 5 y = 18 
16 a: — 3 y = 10 


n f 5 a + 3 6 = —9 
j 3 a — 4 6 = — 17 

12 f8p + 5g = 5 

1 3 p - 2 q = 29 

13 f 6 a; —j— 2 2/= 3 

1 5z - 3y = - 6 

14 | 5 m — 12 n = — 31 
l3m+22n=-4 

15. | 3 « + 7 < = 4 
1 7 s + 8 < = 26 

16. |2* -3y - - 14 
\3z + 7^ =48 


Test XV — Quadratic Equations 
Time: 15 Minutes 

Solve the odd ones by the factoring method; the even ones 


by completing the square 
directs. 

1. z 2 - 36 = 0 

2. z 2 —5z + 6= 0 

3. x 2 + 2 x - 24 = 0 

4. 2 z 2 — 11 x + 5 = 0 
6. 2 w 2 + 9 m + 4 = 0 

6. 2y 2 — 3y—5=0 

7. 3 m 2 + 4 m - 7 = 0 

8. r 2 = 144 

9. x 2 = 9 # — 20 

10. ?/ 2 — 4 y = 45 

11. 8 a 2 = 22 a — 15 

12. 12 t 2 + 13 t + 3 = 0 


by the formula as your teacher 

13. 9 r 2 - 6 r = 8 

14. 2 a 2 = 15 — x 

15. 4 r 2 - 25 = 0 

16. x 2 + 35 = 12 x 

17. y 2 = 7 y -f 18 

18. 5 c 2 + 12 = 23 c 

19. 2 m 2 — 3m =20 

20. 15 c 2 = 4 c + 3 

21. x 2 = 21 -4 a: 

22. 3 a: 2 = 10 - x 

23. 6 2/ 2 — 7 2/ = 3 

24. 8* 2 + 2* = 15 


460 


ALGEBRA 


DEFINITIONS, RULES, AND PROCESSES 

YOU SHOULD KNOW NOW 


I. Definitions. — Be able to give the meaning of each of the following 
words or phrases, either as given in the book (which is preferable, when 
a definition is given in the book), or in your own words. 


1. Literal number 

2. Product 

3. Formula 

4. Equation 

5. Factors of a number 

6. Ray 

7. Angle 

8. Acute angle 

9. Obtuse angle 

10. Complementary angles 

11. Complement 

12. Supplementary angles 

13. Supplement 

14. Straight angle 

15. Perimeter 

16. Algebraic expression 

17. Power of a number 

18. Base (of a power) 

19. Exponent 

20. Absolute value 

21. Monomial or term 

22. Coefficient 

23. Like terms 

24. Unlike terms 

25. Polynomial 

26. Binomial 

27. Trinomial 

28. Ascending powers 

29. Descending powers 

30. Sum 

31. Minuend 

32. Subtrahend 

33. Remainder 


34. Parentheses 

35. Brackets 

36. Identity 

37. Multiplier 

38. Multiplicand 

39. Product 

40. Conditional equation 

41. First degree equation 

42. To factor a number 

43. Prime number 

44. Square root of a number 

45. Quadratic equation, or second 

degree equation 

46. Root of an equation 

47. Literal equation 

48. Common factor 

49. Numerator 

50. Denominator 

51. Multiple 

52. Common multiple 

53. Lowest common multiple 

54. Integral expression 

55. Mixed expression 

56. Complex fraction 

57. Ratio 

58. Proportion 

59. Variable 

60. Horizontal axis 

61. Vertical axis 

62. Origin 

63. Abscissa 

64. Ordinate 

65. Coordinates 


OUTLINE FOR REVIEWS 


461 


66. Independent equations 

78. 

Imaginary unit 

67. Dependent equations 

79. 

Rational number 

68. Simultaneous equations 

80. 

Irrational number 

69. Inconsistent equations 

81. 

Irrational equation 

70. Surd 

82. 

Logarithm 

71. Pure quadratic equation 

83. 

Mantissa 

72. Complete quadratic 

84. 

Characteristic 

73. Integral polynomial 

85. 

Arithmetic progression 

74. Rational polynomial 

86. 

Arithmetic means 

75. Conjugate surds 

87. 

Geometric progression 

76. Imaginary number 

88. 

Geometric means 

77. Complex number 

89. 

Binomial theorem 


II. Rules and facts you should be able to give. 

1. Formula for the area of 

a. a rectangle; 

b. a parallelogram; 

c. a triangle; 

d. a trapezoid; 

e. a circle. 

2. Formula for the length or circumference of a circle. 

3. Formula for the volume of a rectangular parallelopiped. 

4. Four fundamental" rules used in solving a simple equation. 

5. What is the sum of the angles of a triangle? 

6. What is the sum of all the angles around a point? 

7. The order in which fundamental operations are to be performed. 

(§ 22 ) 

8. Rule for adding numbers having 

a. like signs; b. unlike signs. 

9. Rule for multiplying numbers having: 
a. like signs; b. unlike signs. 

10. Rule for adding like terms. 

11. Rule for subtraction in algebra. 

12. Rules for removing parentheses. 

13. Law of exponents for multiplication. 

14. Law of exponents for division. 

15. Two fundamental principles of fractions. 

16. Rule for adding or subtracting fractions. 

17. Rule for multiplying fractions. 

18. Rule for dividing by a fraction. 





462 


ALGEBRA 


III. Examples and problems you should be able to solve and 
check. 

1. How to add polynomials. 

2. How to subtract one polynomial from another. 

3. How to multiply one polynomial by another. 

4. How to divide one polynomial by another. 

5. How to multiply mentally binomials of the form ax + b, as taught 

on page 128. 

6. How to factor a polynomial of the form ax 2 + bx + c, as taught 

on page 130. 

7. How to add and subtract fractions. 

8. How to multiply and divide fractions. 

9. How to solve a simple equation like the ones on page 107. 

10. How to solve a fractional equation like: 

a. numbers 1-16, on page 187; 

b. numbers 18-27, on page 188; 

c. numbers 29-54, on page 188. 

11. How to draw the graph of a simple equation, as on page 216. 

12. How to solve a system of two linear equations graphically, as on 

page 221. 

13. How to solve an equation by elimination: 

a. by addition or subtraction, as on page 226; 

b. by substitution, as on page 229. 

14. How to solve simultaneous fractional equations, as on page 231. 

15. How to get the square root of a number: 

a. by factoring, as on page 248; 

b. by long division, as on pages 252-254. 

16. How to solve a pure quadratic, as on page 261. 

17. How to solve a complete quadratic : 

a. by factoring, as on pages 148 and 150; 

b. by completing the square, as on page 266; 

c. by formula, as on page 271. 

18. How to form the equation for: 

a. a “number relations” problem; 

b. an “age” problem; 

c. a “distance, rate, and time” problem; 

d. a “money” problem; 

e. a “river” problem; 

/. an “interest” problem; 
g. a “digits” problem; 



OUTLINE FOR REVIEWS 


463 


h. an “area” problem; 

i. a “ perimeter ’ ’ problem ; 

j. problems about angles; 

k. problems about fractions. 

19. How to factor x 2 — y 2 + 2 yz — z 2 . (291) 

20. How to factor z 4 -f x 2 y 2 + y*. (292) 

21. How to factor by grouping. (293) 

22. How to factor x n ± y n . (294) 

23. How to factor by the factor theorem. (301) 

24. How to draw the graph of a quadratic equation having two un¬ 

knowns. (309) 

25. How to solve graphically a system involving one or two quad¬ 

ratics. (313) 

26. How to solve by elimination a system consisting: 

a. of a linear and a quadratic; (316) 

b. of two quadratics. (317) 

27. How to tell the kind of roots a given quadratic has, by examina¬ 

tion of the discriminant. 

28. How to find the sum and the product of the roots of a given 

quadratic. 

29. How to reduce a surd to its simplest form. (346) 

30. How to change a surd to an equivalent surd of different order. 

(348) 

31. How to multiply by (349) or to divide by (351-353) a surd. 

32. How to solve an irrational equation. (359) 

33. How to find the logarithm of a number (369) and to find the 

number when its logarithm is given. (372-374) 

34. How to find the nth term and the sum of the first n terms of: 

a. an arithmetic progression; (381, 383) 

b. a geometric progression. (388, 389) 

35. How to expand a power of a binomial by the binomial theorem. 

(397) 




INDEX 


l 




Abscissa, 212. 

Absolute value, 35. 

Addend, 18. 

Algebraic expression, 26; value of 
an, 26. 

Angle, 22; right, 22; straight, 24. 

Angles, complementary, 23; sum 
of, in a triangle, 25; supple¬ 
mentary, 24. 

Arithmetic, mean, 382; progres¬ 
sion, 380. 

Ascending powers, 45. 

Axis, horizontal, 211; vertical, 

211 . 

Base, 26. 

Binomial, 44; square of a, 133; 
theorem, 397. 

Braces, 63. 

Brackets, 63. 

Cancellation, in an equation, 105; 
in a fraction, 157. 

Changing signs, in an equation, 
106 ; in a fraction, 181. 

Characteristic, 365. 

Circumference, 7. 

Clearing of fractions, 186. 

Coefficient, 41; numerical, 41. 

Common, difference, 380; log¬ 
arithm, 364. 

Complement of an angle, 23. 

Complementary angles, 23. 

Complex, fraction, 179; number, 
329. 


Conditional equation, 103. 

Coordinates, 214. 

Degree of an equation, 104. 

Descending powers, 45. 

Difference, 49. 

Digit, 238. 

Discriminant, 330. 

Dividend, 93. 

Division, synthetic, 299. 

Divisor, 93. 

Elimination, by addition or sub¬ 
traction, 226; by substitution, 
229. 

Ellipse, 310. 

Equation, 10; canceling terms in 
an, 105; changing signs in an, 
107; complete quadratic, 265; 
conditional, 103; graphical so¬ 
lution of an, 282; identical, 
103; irrational, 359; linear, 
225; members of an, 10; of 
first degree, 104; properties of 
an, 104; pure quadratic, 260; 
quadratic, 148;' simple, 104; 
solving an, 10; transposition 
in an, 104. 

Equations, dependent, 233; in¬ 
consistent, 233; independent, 
226 ; simultaneous, 226; system 
of, 226. 

Exponent, 26; fractional, 336; 
negative, 338; zero, 337. 


464 



INDEX 


465 


Exponents, law of division of, 95; 
law of multiplication of, 74; 
laws of, 332. 

Expression, algebraic, 26; in¬ 
tegral, 176; mixed, 176. 


Factor, 2, 41; common, 18; 

highest common, 154; to, 118; 
theorem, 301. 

Factors, prime, 127. 

Formula, 3. 

Formulae, deriving, 4, 28, 61, 200. 

Fractions,'156; clearing of, 186; 
equivalent, 169. 

Fulcrum, 194. 

Fundamental operations, 26. 

Geometric, mean, 391; progres¬ 
sion, 388. 

Graph of equation with two varia¬ 
bles, 216. 

Graphical representation, 208. 

Graphical solution of equations 
with one variable, 282. 

Graphs, 209. 

Grouping, symbols of, 63; factor¬ 
ing by, 293. 

Homogeneous equations, 317. 

" Horizontal axis, 214. 

Hyperbola, 312. 

Identity, 103. 

1 Imaginary number, 285. 

J Imaginary numbers, 355. 

! V Imaginary roots of a quadratic 
equation, 285; meaning of, on 
graph,. 287. 

Imaginary unit, 285. 

Inconsistent equations, 233. 

Independent equations, 226. 

Index, 334. 

Infinite geometric progression, 393. 

Integral polynomial, 299. 

Irrational equation, 359; number, 
329. 


Left member of an equation, 10. 

Lever, 194. 

Like terms, 41. 

Linear equation, 225. 

Literal equation, 198. 

Literal number, 1. 

Lowest common multiple, 165. 

Mantissa, 365. 

Members of an equation, 10. 

Minuend, 49. 

Monomial, 41; addition of, 41; 
division of, 96; multiplication 
of, 75; square of, 118; square 
root of, 119. 

Negative numbers, 34; addition 
of, 34; division of, 94; multi¬ 
plication of, 37; powers of, 40; 
subtraction of, 51. 

Negative, term, 41^ exponent, 338. 

Number, imaginary, 285; literal, 
1; negative, 34; positive, 34; 
prime, 118; rational, 329; real, 
329; unknown, 10. 

Numerator, 156. 

Numerical coefficient, 41. 

Numerical value, 41. 

Opposite quantities, 31. 

Ordinate, 214. 

Origin, 214. 

Parabola, 309. 

Parallelogram, 5. 

Parentheses, 8; inclosing terms 
in, 66; removing, 63. 

Periods, 251. 

Polynomial, 44; arranging a, 45. 

Polynomials, addition of, 45; di¬ 
vision of, 99; factoring, 120; 
factoring, by grouping, 293; 
multiplication of, 80; square 
root of, 250; subtraction of, 53. 

Positive, number, 34; quantity, 
32; term, 41. 





466 


INDEX 


Power, 26. 

Powers, ascending, 45; descend¬ 
ing, 45. 

Prime number, 118. 

Product, 2. 

Progression, arithmetic, 380; geo¬ 
metric, 388. 

Proportion, 204. 

Pure quadratic, 260. 

Pyramid, 6. 

Quadratic equation, 148; com¬ 
plete, 265; graph of, 287; hav¬ 
ing two unknowns, 279; im¬ 
aginary roots of, 285; pure, 
260; solution of, by completing 
the square, 265; solution of, by 
factoring, 148; solution of, by 
formula, 270. 

Quadratic surd, 256. 

Quantities, opposite, 31. 

Quantity, negative, 32; positive, 
32. 

Quotient, 93. 

Radical, 343; index of, 334; order 
of, 343. 

Radicals, similar, 347. 

Radicand, 334. 

Ratio, 203; of a geometric pro¬ 
gression, 388. 

Rational number, 329; poly¬ 
nomial, 299. 

Rationalizing the denominator, 357. 

Remainder, 49; theorem, 298. 

Right angle, 22. 

Right member, 10. 

Right triangle, 262. 

Root, cube, 334; of an equation, 
10; principal, 334; square, 119. 

Roots, imaginary, of a quadratic, 
285. 


Signs, change of, in an equation, 
106; law of, in addition, 36; 
in a fraction, 181; in division, 
94; in multiplication, 38. 

Simultaneous equations, 226. 

Solution of simultaneous equations, 
225. 

Square root, approximate, 254; 
by inspection, 248; of a mono¬ 
mial, 119; of a number, 251; 
of a polynomial, 250. 

Straight angle, 24. 

Subtraction, 49. 

Subtrahend, 49. 

Sum, 18. 

Supplement, 24. 

Supplementary angles, 24. 

Surd, conjugate, 353; quadratic, 
256 ; addition of, 257. 

Symbols of grouping, 63. 

Synthetic division, 299. 

System of equations, 226, 316. 

Table of square roots, 254 ; of 
logarithms, 367. 

Term, 41; negative, 41; positive, 
41. 

Terms, like, 41; unlike, 41. 

Theorem, binomial, 397; factor, 
301; remainder, 298. 

Transposition, 104. 

Triangle, altitude of, 5; area of, 
5; base of, 5. 

Trinomial, 44. 

Unit, imaginary, 285. 

Unknown number, 10. 

Unlike terms, 41. 

Variables, 225. 

Vertical axis, 214. 

Vinculum, 63. 



























































































